All Solutions
Page 220: Assessment
$$
begin{aligned}
Fs cos theta &= -Fd \
cos theta &= -1 \
theta &= cos^{-1} left( -1 right) \
theta &= boxed{180^{circ}}
end{aligned}
$$
theta = 180^{circ}
$$
The direction of the gravitational force is downwards. Based on the figure, the bob swings from a higher elevation to a lower elevation, so the vertical displacement is also downwards. Since the force and displacement are parallel, the work done is **positive**.
The direction of the tension of the string on the bob is along the string (up). The direction of motion is perpendicular to the string. Since the force and displacement are perpendicular, the work done must be **zero**.
The direction of the gravitational force is downwards. Based on the figure, the bob swings from a lower elevation to a higher elevation, so the vertical displacement is upwards. Since the force and displacement are antiparallel, the work done is **negative**.
The direction of the tension of the string on the bob is along the string (up). The direction of motion is perpendicular to the string. Since the force and displacement are perpendicular, the work done must be **zero**.
$$
begin{aligned}
W &= Fd \
&= mgd \
&= left( 0.75~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 0.11~mathrm{m} right) \
&= 0.80933~mathrm{J} \
W &= boxed{ 0.81~mathrm{J} }
end{aligned}
$$
W = 0.81~mathrm{J}
$$
$$
W=Fcdot d,
$$
where height is denoted by $d$.
The weightlifter lifts weights straight up, so the direction of the force and movement are the same.
Solve the equation for $F$ and substitute the values for work and height given in the task:
$$
F=frac{W}{d}=frac{9.8:J}{0.12:m}
$$
$$
boxed{F=81.67:N}
$$
F=81.67:N
$$
$$
begin{aligned}
W &= Fd \
W &= wd \
implies d &= frac{W}{w} \
&= frac{13~mathrm{J}}{35~mathrm{N}} \
&= 0.37143~mathrm{m} \
d &= boxed{0.37~mathrm{m}}
end{aligned}
$$
d = 0.37~mathrm{m}
$$
$$
begin{aligned}
W &= Fd cos theta \
&= left( 5.2~mathrm{N} right) left( 0.45~mathrm{m} right) cos 21^{circ} \
&= 2.18458~mathrm{J} \
W &= boxed{2.2~mathrm{J}}
end{aligned}
$$
W = 2.2~mathrm{J}
$$
For this part, we calculate the work done by the person lifting the can. The work is positive, since the force is upwards and the displacement is also upwards. The force must also be equal to the weight of the can. We have
$$
begin{aligned}
W &= Fd \
&= mgd \
&= left( 3.4~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 1.8~mathrm{m} right) \
&= 60.0372~mathrm{J} \
W &= 60.~mathrm{J}
end{aligned}
$$
For this part, the person holds the can to be stationary. We calculate the work done in this time. The displacement is $0$, so the work must also be **zero**. Hence,
$$
W = 0
$$
For this part, the person returns the can to the ground. The force is still upwards, but the displacement is downward, so the work must be negative. We have
$$
begin{aligned}
W &= -Fd \
&= -mgd \
&= -left( 3.4~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 1.8~mathrm{m} right) \
&= -60.0372~mathrm{J} \
W &= -60.~mathrm{J}
end{aligned}
$$
begin{aligned}
W &= 60.~mathrm{J} \
W &= 0 \
W &= -60.~mathrm{J}
end{aligned}
$$
The skier moves toward the boat. The force of the rope on the skier is also pointing toward the boat. The force and displacement are parallel, so the work must be **positive**.
We calculate the work done. We are given the force and displacement, so we have
$$
begin{aligned}
W &= Fd \
&= left( 120~mathrm{N} right) left( 65~mathrm{m} right) \
W &= boxed{7800~mathrm{J}}
end{aligned}
$$
The tension of the rope on the boat is pointing towards the back of the boat. The displacement, however, is pointing to the front of the boat. The force and displacement are antiparallel, so the work done must be **negative**.
We calculate the work done. We are given the force and displacement, so we have
$$
begin{aligned}
W &= -Fd \
&= -left( 120~mathrm{N} right) left( 65~mathrm{m} right) \
W &= boxed{-7800~mathrm{J}}
end{aligned}
$$
$$
begin{aligned}
W &= Fd cos theta \
&= left(16~mathrm{N}right) left(12~mathrm{m}right) cos 25^{circ} \
&= 174.01110~mathrm{J} \
W &= boxed{170~mathrm{J}}
end{aligned}
$$
W = 170~mathrm{J}
$$
$$
begin{aligned}
W &= Fd cos theta \
&= left(120~mathrm{N}right) left(18~mathrm{m}right) cos 43^{circ} \
&= 1579.72400~mathrm{J} \
W &= boxed{1600~mathrm{J}}
end{aligned}
$$
W = 1600~mathrm{J}
$$
$$
begin{aligned}
W &= Fd cos theta \
cos theta &= frac{W}{Fd} \
implies theta &= cos^{-1} left[ frac{W}{Fd} right] \
&= cos^{-1} left[ frac{2500~mathrm{J}}{left(75~mathrm{N}right) left(42~mathrm{m}right)} right] \
&= 37.47200^{circ} \
theta &= boxed{37^{circ}}
end{aligned}
$$
theta = 37^{circ}
$$
| Jogger | Mass | Speed |
| — | — | — |
| A | $m$ | $v$ |
| B | $m/2$ | $3v$ |
| C | $3m$ | $v/2$ |
| D | $4m$ | $v/2$ |
$$
KE = frac{1}{2} mv^{2}
$$
$$
begin{aligned}
KE_{A} &= frac{1}{2} m_{A}v_{A}^{2} &= left( frac{1}{2} right)mv^{2} \
KE_{B} &= frac{1}{2} m_{B}v_{B}^{2} &= left( frac{9}{4} right)mv^{2} \
KE_{C} &= frac{1}{2} m_{C}v_{C}^{2} &= left( frac{3}{8} right)mv^{2} \
KE_{D} &= frac{1}{2} m_{D}v_{D}^{2} &= left( frac{1}{2} right)mv^{2} \
end{aligned}
$$
$$
boxed{KE_{C} < KE_{A} = KE_{D} < KE_{B}}
$$
KE_{C} < KE_{A} = KE_{D} < KE_{B}
$$
$$
KE = frac{1}{2}mv^{2}
$$
All of the factors in the right hand side of the equation can not be negative, $1/2$ is a positive constant, mass $m$ is always nonnegative, and the square of speed $v^{2}$ is always nonnegative, so the product, the kinetic energy, can not be negative.
Let $v = 50~mathrm{km/h}$. The first part is accelerating from $0$ to $v$, and the second part is accelerating from $v$ to $3v$. From the work-energy theorem, we have
$$
begin{aligned}
W &= Delta KE \
&= frac{1}{2}m left[v_text{f}^{2} – v_text{i}^{2} right] \
W &propto left[v_text{f}^{2} – v_text{i}^{2} right]
end{aligned}
$$
$$
8W
$$
Based on the calculations above, the best explanation is **B.** “the final speed is three times the speed that was produced by the work $W$.”
The change in gravitational potential is dependent only on the mass and change in elevation. Since the balls are of the same mass and start from the same height and end on the ground, the change in gravitational potential energy of Ball 1 is **equal** to that of Ball 2.
Based on the explanation above, the best choice is **B.** “the gravitational potential energy depends only on the mass of the ball and its initial height above the ground.”
Based on the diagram, points C and A are in the same elevation. Their graviatational potential energy must be **equal**.
Based on the diagram, point C is in a higher elevation than point B. The gravitational potential energy at point C is **greater than** that at point B.
$$
begin{aligned}
KE &= frac{1}{2}mv^{2}\
&= frac{1}{2} left(1770~mathrm{kg}right) left(120~mathrm{m/s}right)^{2} \
&= 1.27440 times 10^{7}~mathrm{J} \
KE &= boxed{1.3 times 10^{7}~mathrm{J}}
end{aligned}
$$
KE = 1.3 times 10^{7}~mathrm{J}
$$
$$
begin{aligned}
PE_text{gravity} &= mgh \
&= left(7.3~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(1.7~mathrm{m}right) \
&= 121.74210~mathrm{J} \
PE_text{gravity} &= boxed{120~mathrm{J}}
end{aligned}
$$
PE_text{gravity} = 120~mathrm{J}
$$
$$
begin{aligned}
KE &= frac{1}{2}mv^{2} \
v^{2} &= frac{2KE}{m} \
implies v &= sqrt{ frac{2KE}{m} } \
&= sqrt{frac{2left(18~mathrm{J}right)}{0.15~mathrm{kg}}} \
&= 15.49193~mathrm{m/s} \
v &= boxed{15~mathrm{m/s}}
end{aligned}
$$
v = 15~mathrm{m/s}
$$
$$
begin{aligned}
PE_text{gravity} &= mgh \
implies h &= frac{PE_text{gravity}}{mg} \
&= frac{6.6~mathrm{J}}{left(0.12~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right)} \
&= 5.60652~mathrm{m} \
h &= boxed{5.6~mathrm{m}}
end{aligned}
$$
h = 5.6~mathrm{m}
$$
$$
begin{aligned}
PE_text{spring} &= frac{1}{2}kx^{2} \
&= frac{1}{2}left(92~mathrm{N/m}right) left(2.8 times 10^{-2}~mathrm{m}right)^{2}\
&= 3.60640 times 10^{-2}~mathrm{J} \
PE_text{spring} &= boxed{ 3.6 times 10^{-2}~mathrm{J} }
end{aligned}
$$
PE_text{spring} = 3.6 times 10^{-2}~mathrm{J}
$$
$$
begin{aligned}
F &= kx_{1} \
implies k &= frac{F}{x_{1}}
end{aligned}
$$
$$
begin{aligned}
PE_text{spring} &= frac{1}{2}kx_{2}^{2} \
&= frac{1}{2} frac{F}{x_{1}}x_{2}^{2} \
&= frac{1}{2}frac{27~mathrm{N}}{4.4 times 10^{-2}~mathrm{m}} left(3.5 times 10^{-2}~mathrm{m}right)^{2} \
&= 0.37585~mathrm{J} \
PE_text{spring} &= boxed{0.38~mathrm{J}}
end{aligned}
$$
PE_text{spring} = 0.38~mathrm{J}
$$
$$
begin{aligned}
PE_text{spring} &= frac{1}{2}kx^{2} \
implies k &= frac{2PE_text{spring}}{x^{2}} \
&= frac{2left(0.053~mathrm{J}right)}{left(2.6 times 10^{-2}~mathrm{m}right)^{2}} \
&= 156.80473~mathrm{N/m} \
k &= boxed{160~mathrm{N/m}}
end{aligned}
$$
k = 160~mathrm{N/m}
$$
$$
begin{aligned}
W &= W \
-Fd &= Delta KE = KE_text{f} – KE_text{i} \
-Fd &= frac{1}{2}mleft[v_text{f}^{2} – v_text{i}^{2}right] \
implies F &= frac{1}{2}frac{m}{d} left[v_text{i}^{2} – v_text{f}^{2}right] \
&= frac{1}{2} frac{1100~mathrm{kg}}{32~mathrm{m}} left[ left(19~mathrm{m/s}right)^{2} – left(12~mathrm{m/s}right)^{2} right] \
&= 3729.68750~mathrm{N} \
F &= boxed{3700~mathrm{N}}
end{aligned}
$$
F = 3700~mathrm{N}
$$
Since the biker and rider stops, the final kinetic energy is $0$. Using the work-energy theorem, we have
$$
begin{aligned}
W &= Delta KE = 0 – KE_text{i} \
&= -frac{1}{2}left(m_{1} + m_{2}right)v^{2} \
&= -frac{1}{2} left(65~mathrm{kg} + 8.8~mathrm{kg}right) left(14~mathrm{m/s}right)^{2} \
&= -7232.4~mathrm{J} \
W &= boxed{-7200~mathrm{J}}
end{aligned}
$$
The brakes apply a force opposite to the direction of motion, so the work done is negative and is $W = -Fd$. We have
$$
begin{aligned}
W &= -Fd \
implies F &= -frac{W}{d} \
&= – frac{-7232.4~mathrm{J}}{3.5~mathrm{m}} \
&= 2066.4~mathrm{N} \
F &= boxed{2100~mathrm{N}}
end{aligned}
$$
$$
begin{aligned}
W &= -7200~mathrm{J} \
end{aligned}$$
(b)
$$begin{aligned}
F &= 2100~mathrm{N}
end{aligned}
$$
The final speed is $0$, so the final kinetic energy is also $0$. Using work-energy theorem, we have
$$
begin{aligned}
W &= Delta KE = 0 – KE_text{i} \
&= -frac{1}{2}mv^{2} \
&= -frac{1}{2} left(62~mathrm{kg}right) left(4.5~mathrm{m/s}right)^{2} \
&= -627.75~mathrm{J} \
W &= boxed{-630~mathrm{J}}
end{aligned}
$$
$$
begin{aligned}
W &= -Fd \
W &= -mu_{k} mgd \
implies mu_{k} &= -frac{W}{mgd} \
&= -frac{-627.75~mathrm{J}}{left(62~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(3.4~mathrm{m}right)} \
&= 0.30356 \
mu_{k} &= boxed{0.30}
end{aligned}
$$
begin{aligned}
W &= -630~mathrm{J} \
mu_{k} &= 0.30
end{aligned}
$$
The mass does not change and it can be calculated using the values at time $t = 0$. From the definition of kinetic energy, we have
$$
begin{aligned}
KE_text{i} &= frac{1}{2}mv_text{i}^{2} \
implies m &= frac{2KE_text{i}}{v_text{i}^{2}} \
&= frac{2left(14~mathrm{J}right)}{left(3.5~mathrm{m/s}right)^{2}} \
&= 2.28571~mathrm{kg} \
m &= boxed{2.3~mathrm{kg}}
end{aligned}
$$
For this part, we calculate its kinetic energy when the speed is $v_text{f} = 4.7~mathrm{m/s}$. We have
$$
begin{aligned}
KE_text{f} &= frac{1}{2}mv_text{f}^{2} \
&= frac{1}{2} left( frac{2KE_text{i}}{v_text{i}^{2}} right)v_text{f}^{2} \
&= KE_text{i} left( frac{v_text{f}}{v_text{i}} right)^{2} \
&= left(14~mathrm{J}right) left(frac{4.7~mathrm{m/s}}{3.5~mathrm{m/s}}right)^{2} \
&= 25.24571~mathrm{J} \
KE_text{f} &= boxed{25~mathrm{J}}
end{aligned}
$$
The work-energy theorem tells us that the work done is the change in kinetic energy. We have
$$
begin{aligned}
W &= Delta KE = KE_text{f} – KE_text{i} \
&= 25.24571~mathrm{J} – 14~mathrm{J} \
&= 11.24571~mathrm{J} \
W &= boxed{11~mathrm{J}}
end{aligned}
$$
begin{aligned}
m &= 2.3~mathrm{kg} \
KE_text{f} &= 25~mathrm{J} \
W &= 11~mathrm{J}
end{aligned}
$$
$$
h = lleft(1 – cos thetaright)
$$
$$
begin{aligned}
Delta PE_text{gravity} &= -mgh \
&= -mglleft(1 – cos thetaright) \
&= -left(0.33~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(1.2~mathrm{m}right) left(1 – cos 35^{circ}right) \
&= -0.70255~mathrm{J} \
Delta PE_text{gravity} &= boxed{-0.70~mathrm{J}}
end{aligned}
$$
Delta PE_text{gravity} = -0.70~mathrm{J}
$$
For this part, we calculate the kinetic energy if the potential energy changes to $PE_text{f} = 15~mathrm{J}$. Using conservation of energy, we have
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
implies KE_text{f} &= KE_text{i} + PE_text{i} – PE_text{f} \
&= 10~mathrm{J} + 20~mathrm{J} – 15~mathrm{J} \
KE_text{f} &= boxed{15~mathrm{J}}
end{aligned}
$$
For this part, we calculate the kinetic energy if the potential energy changes to $PE_text{f} = 5~mathrm{J}$. Using conservation of energy, we have
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
implies KE_text{f} &= KE_text{i} + PE_text{i} – PE_text{f} \
&= 10~mathrm{J} + 20~mathrm{J} – 5~mathrm{J} \
KE_text{f} &= boxed{25~mathrm{J}}
end{aligned}
$$
$$
begin{aligned}
KE_text{f} &= 15~mathrm{J} \
end{aligned}$$
(b)
$$begin{aligned}
KE_text{f} &= 25~mathrm{J}
end{aligned}
$$
$$
begin{aligned}
E &= KE_text{i} + PE_text{i} \
&= 10~mathrm{J} + 30~mathrm{J} \
E &= 40~mathrm{J}
end{aligned}
$$
$$
boxed{ PE_text{max} = 40~mathrm{J} }
$$
$$
boxed{ KE_text{max} = 40~mathrm{J} }
$$
begin{aligned}
PE_text{max} &= 40~mathrm{J} \
KE_text{max} &= 40~mathrm{J}
end{aligned}
$$
The change in kinetic energy is equal to the work done by gravity. Since the balls are identical and starts from the same height, their initial potential energy are equal. Also, both go the the ground, so their final potential energy are equal. The work done by gravity will be equal for the two balls, so the change in kinetic energy will be **equal**.
Based on the explanation above, the best choice is **C.** “the change in gravitational potential energy is the same for each bll, which means that the change in kinetic energy must also be the same”.
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
frac{1}{2}mv_text{i}^{2} + 0 &= frac{1}{2}mv_text{f}^{2} + mgh \
v_text{f}^{2} &= v_text{i}^{2} – 2gh \
implies v_text{f} &= sqrt{v_text{i}^{2} – 2gh}
end{aligned}
$$
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
frac{1}{2}mv_text{i}^{2} + 0 &= frac{1}{2}mv_text{f}^{2} + mgh \
h &= frac{v_text{i}^{2} – v_text{i}^{2}}{2g} \
&= frac{left(8.30~mathrm{m/s}right)^{2} – left(7.10~mathrm{m/s}right)^{2}}{2left(9.81~mathrm{m/s^{2}}right)} \
&= 0.94190~mathrm{m} \
h &= boxed{0.942~mathrm{m}}
end{aligned}
$$
h = 0.942~mathrm{m}
$$
For the first part, the initial speed is $0$, so the initial kinetic energy is
$$
KE_text{i} = boxed{0}
$$
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
0 + mgh &= KE_text{f} + 0 \
implies KE_text{f} &= mgh \
&= left(5.76~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(2.00~mathrm{m}right)\
&= 113.0112~mathrm{J} \
KE_text{f} &= boxed{113~mathrm{J}}
end{aligned}
$$
$$
begin{aligned}
Delta KE &= KE_text{f} – KE_text{i} \
&= 113.0112~mathrm{J} – 0 \
Delta KE &= boxed{113~mathrm{J}}
end{aligned}
$$
For this part, the initial kinetic energy must be the final kinetic energy of the previous part. We have
$$
KE_text{i} = boxed{113~mathrm{J}}
$$
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
KE_text{i} + mgh &= KE_text{f} + 0 \
implies KE_text{f} &= KE_text{i}+ mgh \
&= 113.0112~mathrm{J} + left(5.76~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(2.00~mathrm{m}right)\
&= 226.0224~mathrm{J} \
KE_text{f} &= boxed{226~mathrm{J}}
end{aligned}
$$
$$
begin{aligned}
Delta KE &= KE_text{f} – KE_text{i} \
&= 226.0224~mathrm{J} – 113.0112~mathrm{J} \
Delta KE &= boxed{113~mathrm{J}}
end{aligned}
$$
tt{(a) The zero point $h_0=0$ for potential energy will be chosen at the base of the cliff, ignoring the air resistance the system become ideal and we can use the mechanical energy conservation formula:}
$$
$$
begin{align*}
E_i&=E_f \
PE_{i,gravity}+KE_i&=PE_{f,gravity}+KE_f \
mgh+frac{1}{2}mv_{i}^2&=0+frac{1}{2}mv_{f}^2\
v_i&=sqrt{v_{f}^{2}-2gh}\
&=sqrt{(29frac{m}{s})^2-2(9.81frac{m}{s^2})(32m)}\
&=boxed{color{#4257b2}{14.6frac{m}{s}}}
end{align*}
$$
$$
begin{align*}
PE_{i,gravity}+KE_i&=PE_{f,gravity}+KE_f\
mgh+frac{1}{2}mv_{i}^{2}&=mgh_{max}+0\
h_{max}&=h+frac{v_{i}^{2}}{2g}\
&=(32m)+frac{(14.6frac{m}{s})^2}{2(9.81frac{m}{s^2})}\
&=boxed{color{#4257b2}{42.86m}}
end{align*}
$$
tt{(a) $v_i=14.6frac{m}{s}$, (b) $h_{max}=42.86m$}
$$
$$
begin{aligned}
frac{1}{2}kx^{2} &= frac{1}{2}mv^{2} \
x^{2} &= v^{2} frac{m}{k} \
implies x &= vsqrt{frac{m}{k}} \
&= left(2.2~mathrm{m/s}right)sqrt{frac{3.7~mathrm{kg}}{3200~mathrm{N/m}}} \
&= 7.48081~mathrm{cm} \
x &= boxed{7.5~mathrm{cm}}
end{aligned}
$$
x = 7.5~mathrm{cm}
$$
$$
begin{aligned}
frac{1}{2}mv^{2} &= frac{1}{2}kx^{2} \
v^{2} &= x^{2} frac{k}{m} \
implies v &= xsqrt{frac{k}{m}} \
&= left(4.2 times 10^{-2}~mathrm{m}right)sqrt{frac{1400~mathrm{N/m}}{1.3~mathrm{kg}}} \
&= 1.37829~mathrm{m/s} \
v &= boxed{1.4~mathrm{m/s}}
end{aligned}
$$
v = 1.4~mathrm{m/s}
$$
Point A is at an elevation $h$ above point B, with
$$
h = lleft(1 – cos theta right)
$$
Let point B be the reference point. The change in gravinational potential energy must be
$$
begin{aligned}
Delta PE &= PE_text{f} – PE_text{i} = PE_text{f} – 0 \
&= mgl left(1 – cos theta right) \
&= left(0.33~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(1.2~mathrm{m}right) left(1 – cos 35^{circ} right) \
&= 0.702255~mathrm{J} \
Delta PE &= boxed{0.70~mathrm{J}}
end{aligned}
$$
We use conservation of energy to find the speed at point A. We have
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
frac{1}{2}mv_text{i}^{2} + 0 &= frac{1}{2}mv_text{f}^{2} + mglleft(1 – cos theta right) \
v_text{f}^{2} &= v_text{i}^{2} – 2glleft(1 – cos theta right) \
implies v_text{f} &= sqrt{v_text{i}^{2} – 2glleft(1 – cos theta right)} \
&= sqrt{left(2.4~mathrm{m/s}right)^{2} – 2left(9.81~mathrm{m/s^{2}}right) left(1.2~mathrm{m}right) left(1 – cos 35^{circ} right)} \
&= 1.22561~mathrm{m/s} \
v_text{f} &= boxed{1.2~mathrm{m/s}}
end{aligned}
$$
begin{aligned}
Delta PE &= 0.70~mathrm{J} \
v_text{f} &= 1.2~mathrm{m/s}
end{aligned}
$$
From the definition of kinetic energy, we have
$$
begin{aligned}
KE_text{i} &= frac{1}{2}mv_text{i}^{2} \
&= frac{1}{2} left(0.33~mathrm{kg}right) left(2.4~mathrm{m/s}right)^{2} \
&= 0.9504~mathrm{J} \
KE_text{i} &= boxed{0.95~mathrm{J}}
end{aligned}
$$
For this part, we calculate the change in gravitational potential energy when the bob moves from point B to the point it hs zero speed. Since the speed is zero, all of htei nitial kinetic energy must have been converted into potential energy. The change in potential energy must be equal to the kinetic energy when the bob is at point B. We have
$$
Delta PE = 0.95~mathrm{J}
$$
We find the angle corresponding to the point in which the speed is zero. We have, by conservation of energy,
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
frac{1}{2}mv_text{i}^{2} + 0 &= 0 + mglleft(1 – cos theta right) \
1 – cos theta &= frac{v_text{i}^{2}}{2gl} \
cos theta &= 1 – frac{v_text{i}^{2}}{2gl} \
implies theta &= cos^{-1} left[ 1 – frac{v_text{i}^{2}}{2gl} right] \
&= 40.94389^{circ} \
theta &= boxed{41^{circ}}
end{aligned}
$$
begin{aligned}
KE_text{i} &= 0.95~mathrm{J} \
Delta PE &= 0.95~mathrm{J} \
theta &= 41^{circ}
end{aligned}
$$
For this part, we let their initial position be the position of $0$ potential energy. Since the masses are connected by a fixed length rope, their speeds must always be equal. Using energy conservation, we have
$$
begin{aligned}
KE_text{i} + PE_text{i} &= PE_text{f} + KE_text{f} \
0 + 0 &= m_{1}gh + m_{2}g left(-hright) + frac{1}{2}left(m_{1} + m_{2}right)v^{2} \
v^{2} &= frac{2gh left(m_{2} – m_{1}right)}{m_{2} + m_{1}} \
implies v &= boxed{ sqrt{frac{2gh left(m_{2} – m_{1}right)}{m_{2} + m_{1}}} }
end{aligned}
$$
Substituting the given quantitles into the expression we have for $v$, we get
$$
begin{aligned}
implies v &= sqrt{frac{2gh left(m_{2} – m_{1}right)}{m_{2} + m_{1}} } \
&= sqrt{frac{2left(9.81~mathrm{m/s^{2}}right) left(1.2~mathrm{m}right) left(4.1~mathrm{kg} – 3.7~mathrm{kg}right)}{4.1~mathrm{kg} + 3.7~mathrm{kg}}} \
&= 1.09881~mathrm{m/s} \
v &= boxed{1.1~mathrm{m/s}}
end{aligned}
$$
begin{aligned}
v &= sqrt{frac{2ghleft(m_{2}-m_{1}right)}{m_{2} + m_{1}}} \
v &= 1.1~mathrm{m/s}
end{aligned}
$$
$$
begin{aligned}
P &= frac{W}{t} \
implies W &= Pt
end{aligned}
$$
$$
begin{aligned}
W_{1} &= W_{2} \
P_{1}t_{1} &= P_{2}t_{2} \
left(2P_{2}right)T &= P_{2}t_{2} \
implies t_{2} &= boxed{2T}
end{aligned}
$$
Since engine 2 has half the power of engine 1, it would take twice the time to product the same power.
$$
begin{aligned}
P_{1} &= frac{W}{T}
end{aligned}
$$
$$
begin{aligned}
P_{2} &= frac{P_{1}}{2} \
P_{2} &= frac{W}{2T}
end{aligned}
$$
$$
begin{aligned}
P_{2} &= frac{3W}{t_{2}} \
implies t_{2} &= frac{3W}{W/2T} \
t_{2} &= boxed{6T}
end{aligned}
$$
| Force | Work | Time |
| — | — | — |
| A | $5~mathrm{J}$ | $10~mathrm{s}$ |
| B | $3~mathrm{J}$ | $5~mathrm{s}$ |
| C | $6~mathrm{J}$ | $18~mathrm{s}$ |
| D | $25~mathrm{J}$ | $125~mathrm{s}$ |
We rank the forces in order of increasing power produced.
$$
P = frac{W}{t}
$$
$$
begin{aligned}
P_{A} &= frac{W_{A}}{t_{A}} = frac{5~mathrm{J}}{10~mathrm{s}} &= 0.5~mathrm{W} \
P_{B} &= frac{W_{B}}{t_{B}} = frac{3~mathrm{J}}{5~mathrm{s}} &= 0.6~mathrm{W} \
P_{C} &= frac{W_{C}}{t_{C}} = frac{6~mathrm{J}}{18~mathrm{s}} &= 0.3~mathrm{W} \
P_{D} &= frac{W_{D}}{t_{D}} = frac{25~mathrm{J}}{125~mathrm{s}} &= 0.2~mathrm{W} \
end{aligned}
$$
$$
boxed{P_{D} < P_{C} < P_{A} < P_{B}}
$$
P_{D} < P_{C} < P_{A} < P_{B}
$$
| Force | Work | Power |
| — | — | — |
| A | $40~mathrm{J}$ | $80~mathrm{W}$ |
| B | $35~mathrm{J}$ | $5~mathrm{W}$ |
| C | $75~mathrm{J}$ | $25~mathrm{W}$ |
| D | $60~mathrm{J}$ | $30~mathrm{W}$ |
We rank the forces in increasing time required to do the work.
$$
begin{aligned}
P &= frac{W}{t} \
implies t &= frac{W}{P}
end{aligned}
$$
$$
begin{aligned}
t_{A} &= frac{W_{A}}{P_{A}} = frac{40~mathrm{J}}{80~mathrm{W}} &=0.5~mathrm{s} \
t_{B} &= frac{W_{B}}{P_{B}} = frac{35~mathrm{J}}{5~mathrm{W}} &=7~mathrm{s} \
t_{C} &= frac{W_{C}}{P_{C}} = frac{75~mathrm{J}}{25~mathrm{W}} &=3~mathrm{s} \
t_{D} &= frac{W_{D}}{P_{D}} = frac{60~mathrm{J}}{30~mathrm{W}} &=2~mathrm{s}
end{aligned}
$$
$$
boxed{ t_{A} < t_{D} < t_{C} < t_{B} }
$$
t_{A} < t_{D} < t_{C} < t_{B}
$$
$$
P = 1000~mathrm{W}
$$
$$
t = 3600~mathrm{s}
$$
$$
begin{aligned}
E &= Pt \
&= left(1000~mathrm{W}right) left(3600~mathrm{s}right) \
&= 3600000~mathrm{J} \
E &= boxed{3.6 times 10^{6}~mathrm{J}}
end{aligned}
$$
E = 3.6 times 10^{6}~mathrm{J}
$$
$$
begin{aligned}
P &= frac{W}{t} \
&= frac{mgh}{t} \
&= frac{left(65~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(3.0~mathrm{m}right)}{12~mathrm{s}} \
&= 159.4125~mathrm{W} \
P &= boxed{160~mathrm{W}}
end{aligned}
$$
$$
begin{aligned}
P &= 159.4125~mathrm{W} frac{1~mathrm{hp}}{746~mathrm{W}} \
&= 0.21369~mathrm{hp} \
P &= boxed{0.21~mathrm{hp}}
end{aligned}
$$
P = 160~mathrm{W} = 0.21~mathrm{hp}
$$
$$
begin{aligned}
P &= Fv \
implies v &= frac{P}{F} \
&= frac{P}{w} \
&= frac{72~mathrm{W}}{88~mathrm{N}} \
&= 0.81818~mathrm{m/s} \
v &= boxed{0.82~mathrm{m/s}}
end{aligned}
$$
v = 0.82~mathrm{m/s}
$$
$$
begin{aligned}
P &= frac{W}{t} \
&= frac{mgh}{t} \
&= ghfrac{m}{t} \
&= left(9.81~mathrm{m/s^{2}}right) left(2.1~mathrm{m}right) left(12~mathrm{kg/s}right) left( frac{1~mathrm{hp}}{746~mathrm{W}}right) \
&= 0.33138~mathrm{hp} \
P &= boxed{0.33~mathrm{hp}}
end{aligned}
$$
P = 0.33~mathrm{hp}
$$
We first convert the given power and time into SI units
$$
begin{aligned}
P &= 0.30~mathrm{hp} times frac{746~mathrm{W}}{1~mathrm{hp}} = 223.8~mathrm{W} \
t &= 2~mathrm{hour}~49~mathrm{min} = 10140~mathrm{s}
end{aligned}
$$
$$
begin{aligned}
E &= Pt \
&= left(223.8~mathrm{W}right) left(10140~mathrm{s}right) \
&= 2269332~mathrm{J} \
E &= 2.3 times 10^{6}~mathrm{J}
end{aligned}
$$
The amount of energy per candy bar must be
$$
begin{aligned}
E_text{bar} &= 280~mathrm{Cal} times frac{4186~mathrm{J}}{1~mathrm{Cal}} = 1172080~mathrm{J}
end{aligned}
$$
$$
begin{aligned}
frac{E}{E_text{bar}} &= frac{2269332~mathrm{J}}{1172080~mathrm{J}} \
&= 1.93626 \
&= boxed{2}
end{aligned}
$$
We first convert the time into SI units. We have
$$
t = 3.25~mathrm{days} = 280800~mathrm{s}
$$
$$
begin{aligned}
P &= frac{W}{t} \
&= frac{mgh}{t} \
&= frac{left(4.35~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(0.760~mathrm{m}right)}{280800~mathrm{s}} \
&= 1.15498 times 10^{-4}~mathrm{W} \
P &= boxed{1.15 times 10^{-4}~mathrm{W}}
end{aligned}
$$
From the definition of power, we have
$$
begin{aligned}
P &= frac{W}{t} \
P &propto frac{1}{t}
end{aligned}
$$
The relationship between the power delivered and time is inverse. To increase the power, the time must be **decreased.**
The potential energy of the ball must start from its maximum value to its minimum, since the potential energy decreases as the ball falls. The graph of potential energy is **graph B**.
The kinetic energy starts from 0, since the ball starts from rest, and continues increasing until the ball hits the ground. The graph of kinetic energy is **graph C.**
The total mechanical energy is constant, and **graph A** is the graph corresponding to a constant.
$$
begin{aligned}
PE &= frac{1}{2}kx^{2} \
&= frac{1}{2}left(310~mathrm{N/m}right)x^{2} \
PE &= boxed{ left(155~mathrm{N/m} right)x^{2} }
end{aligned}
$$
| $x$ | $PE$ |
| — | — |
| $1.0~mathrm{cm}$ | $0.0155~mathrm{J}$ |
| $2.0~mathrm{cm}$ | $0.062~mathrm{J}$ |
| $3.0~mathrm{cm}$ | $0.1395~mathrm{J}$ |
| $4.0~mathrm{cm}$ | $0.248~mathrm{J}$ |
$$
v = frac{h}{t}
$$
$$
F = w
$$
$$
begin{aligned}
P &= Fv \
&= left(wright) left(frac{h}{t}right) \
&= left(836~mathrm{N}right) left(frac{10.7~mathrm{m}}{23.2~mathrm{s}}right)\
&= 385.56897~mathrm{W} \
P &= boxed{386~mathrm{W}}
end{aligned}
$$
P = 386~mathrm{W}
$$
$$
begin{aligned}
E &= 280~mathrm{Cal} times frac{4186~mathrm{J}}{1~mathrm{Cal}} \
E &= 1172080~mathrm{J}
end{aligned}
$$
$$
begin{aligned}
P &= frac{E}{t} \
implies t &= frac{E}{P} \
&= frac{1172080~mathrm{J}}{22~mathrm{W}} \
&= 53276.36364~mathrm{s} \
t &= boxed{5.3 times 10^{5}~mathrm{s}}
end{aligned}
$$
t = 5.3 times 10^{5}~mathrm{s}
$$
$Rightarrowquad$Power of $1$ Atmos cloch $=Big(dfrac{60.0}{240times10^6}Big)W$
So, energy required to run an Atmos clock for $1$ day ,
$E=(Ptimes t)$
$Rightarrowquad E=Big[Big(dfrac{60.0}{240times10^6}Big)times(1times24times3600)Big]J$
$$
Rightarrowquadboxed{E=(0.0216)J}
$$
$$
begin{aligned}
0 &= F – mu_{k}mg \
implies F &= mu_{k} mg
end{aligned}
$$
$$
begin{aligned}
P &= Fv \
&= mu_{k} mgv \
&= left(0.55right) left(67~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(0.50~mathrm{m/s}right) \
&= 180.74925~mathrm{W} \
P &= boxed{180~mathrm{W}}
end{aligned}
$$
P = 180~mathrm{W}
$$
For this part, the mop is angled $theta = 55^{circ}$ above the horizontal, and the mop is pushed for $d = 0.50~mathrm{m}$. We calculate the work done.
$$
begin{aligned}
W &= Fd cos theta \
&= left(43~mathrm{N}right) left(0.50~mathrm{m}right) cos 55^{circ} \
&= 12.33189~mathrm{J} \
W &= boxed{12~mathrm{J}}
end{aligned}
$$
For this part, the angle is increased to $theta = 65^{circ}$. We calculate what happens to the work done.
$$
begin{aligned}
W &= Fd cos theta \
cos theta &= frac{W}{Fd} \
implies theta &= cos^{-1} left[ frac{W}{Fd} right] \
&= cos^{-1} left[ frac{2.00 times 10^{5}~mathrm{J}}{left(2560~mathrm{N}right) left(145~mathrm{m}right)} right] \
&= 57.39848^{circ} \
theta &= boxed{57.4^{circ}}
end{aligned}
$$
theta = 57.4^{circ}
$$
For this part, we know the relationship between the power produced, force, and the speed
$$
begin{aligned}
P &= Fv \
&= left(21~mathrm{N}right) left(0.23~mathrm{m/s}right) \
&= 4.83~mathrm{W} \
P &= boxed{4.8~mathrm{W}}
end{aligned}
$$
First, we convert the time given into SI units
$$
begin{aligned}
t &= 1.5~mathrm{min} times frac{60~mathrm{s}}{1~mathrm{min}} \
t &= 90~mathrm{s}
end{aligned}
$$
$$
begin{aligned}
W &= Pt \
&= left(4.83~mathrm{W}right) left(90~mathrm{s}right) \
&= 434.7~mathrm{J} \
W &= boxed{430~mathrm{J}}
end{aligned}
$$
begin{aligned}
P &= 4.8~mathrm{W} \
W &= 430~mathrm{J}
end{aligned}
$$
$B:qquad KE_B=(0)Jqquad,qquad PE_B=(25)J$
$C:qquad KE_C=?qquad,qquad PE_C=(5)J$
due to absence of friction, total mechanical energy in conserved.
$a)quad KE_A+PE_A=KE_B+PE_B$
$Rightarrowquad 12+PE_A=0+25quadRightarrowquadboxed{PE_A=(13)J}$
$b)quad KE_B+PE_B=KE_C+PE_C$
$$
Rightarrowquad 0+25=KE_C+5quadRightarrowquadboxed{KE_C=(20)J}
$$
Right before the ball falls, all of its initial potential energy is converted to kinetic energy. Using energy conservation, we have
$$
KE = mgh
$$
$$
begin{aligned}
KE_{2} &= left(4mright)gleft(h/4right) \
KE_{2} &= mgh \
KE_{2} &= boxed{KE}
end{aligned}
$$
We see that despite having different masses and initial heights, both of the balls have equal initial potential energy, so their final kinetic energy must also be equal.
Based on the solution above, the best solution is **A.** “the two balls have the same initial energy.”
$$
begin{aligned}
W &= Delta KE = KE_text{f} – KE_text{i} \
-Fd &= 0 – frac{1}{2}mv^{2} \
implies F &= frac{1}{2} frac{m}{d} v^{2} \
&= frac{1}{2} frac{12~mathrm{kg}}{0.22~mathrm{m}} left(550~mathrm{m/s}right)^{2} \
&= 8.25 times 10^{6}~mathrm{N} \
F &= boxed{8.2 times 10^{6}~mathrm{N}}
end{aligned}
$$
F = 8.2 times 10^{6}~mathrm{N}
$$
$m=(3.6)kg$
power output,$quad P=(22)W$
Now,$quad P=Fv=(mg)v$
$Rightarrowquad v=dfrac{P}{mg}=dfrac{(22)W}{(3.6)(9.8)N}$
$Rightarrowquadunderline{v=(0.62)m/s}$
So, the $3.6$ $kg$ milk container must be lifted at rate of $boxed{(0.62)m/s}$ for power output of arm to be $(22)W$
$b)quad$resistance,$quad d=(1.0)m$
so,$quad d=vt$
$Rightarrowquad$Time $(t)=(d/v)=(dfrac{1.0}{0.62})sapprox(1.6)s$
Hence, it takes $boxed{(1.6)s}$ to lift the container through a distance of $(1.0)m$
The work done must be equal to the change in kinetic energy of the jet
$$
begin{aligned}
W &= Delta KE = KE_text{f} – KE_text{i} \
W &= frac{1}{2}mv_text{f}^{2} – 0 \
implies m &= frac{2W}{v_text{f}^{2}} \
&= frac{2left(7.6 times 10^{7}~mathrm{J}right)}{left(72~mathrm{m/s}right)^{2}} \
&= 2.93210 times 10^{4}~mathrm{kg} \
m &= boxed{2.9 times 10^{4}~mathrm{kg}}
end{aligned}
$$
The power output is the rate in which work is done, so we have
$$
begin{aligned}
P &= frac{W}{t} \
&= frac{7.6 times 10^{7}~mathrm{J}}{2.0~mathrm{s}} \
P &= boxed{3.8 times 10^{7}~mathrm{W}}
end{aligned}
$$
begin{aligned}
m &= 2.9 times 10^{4}~mathrm{kg} \
P &= 3.8 times 10^{7}~mathrm{W}
end{aligned}
$$
The magnitude of displacement is equal to the speed times time. We have
$$
d = vt
$$
$$
begin{aligned}
W_text{rope} &= Fd cos theta \
&= Fvt cos theta \
&= left(90.0~mathrm{N}right) left(14~mathrm{m/s}right) left(10.0~mathrm{s}right) cos 35^circ \
&= 1.03213 times 10^{4}~mathrm{J} \
W_text{rope} &= boxed{1.0 times 10^{4}~mathrm{J}}
end{aligned}
$$
The total work done on the object is equal to the change in kinetic energy. Since the speed is constant, the kinetic energy must be contant and the total work done must be zero. To achieve this, the work done by the water must be equal in magnitude to the work done by the rope but opposite in sign. Hence
$$
W_text{water} = boxed{-1.0 times 10^{4}~mathrm{J}}
$$
begin{aligned}
W_text{rope} &= 1.0 times 10^{4}~mathrm{J} \
W_text{water} &= -1.0 times 10^{4}~mathrm{J}
end{aligned}
$$
Velocity of spider , $v=(2.3)cm/s=(2.3times10^{-2})m/s$
$theta=25^circ$
So , power output of spider :
$P=|vec Fcdotvec v|=big((mgcostheta)(v)big)$
$Rightarrowquad P=[(1.8times10^{-3})(9.8)(cos25^circ)times(2.3times10^{-2})]W$
$$
Rightarrowquadboxed{P=(3.7times10^{-4})W}
$$
The power produced is equal to the rate in which work is done, and we have
$$
begin{aligned}
P &= frac{W}{t} \
&= frac{45.5175~mathrm{J}}{0.075~mathrm{s}} \
&= 606.9~mathrm{W} \
P &= boxed{610~mathrm{W}}
end{aligned}
$$
From the definition of power, we have
$$
begin{aligned}
P &= frac{W}{t} \
implies P &propto frac{1}{t}
end{aligned}
$$
The relationship between the power produced and the time it does work is inverse proportion. Since the time is decreased, the power produce must be **more than** the power we initially calculated.
First, we convert the given time into SI units
$$
begin{aligned}
t &= 1~mathrm{day} times frac{24~mathrm{hours}}{1~mathrm{day}} times frac{3600~mathrm{s}}{1~mathrm{hour}} \
t &= 86400~mathrm{s}
end{aligned}
$$
$$
begin{aligned}
E &= Pt \
&= left(1.33~mathrm{W}right) left(86400~mathrm{s}right) \
&= 1.14912 times 10^{5}~mathrm{J} \
E &= boxed{1.15 times 10^{5}~mathrm{J}}
end{aligned}
$$
The amount of energy used in climbing a set of stairs must be equal to the gravitational potential energy at the top step with total height $h$. An average person has mass $m = 65.0~mathrm{kg}$ and we use $g = 9.81~mathrm{m/s^{2}}$.
$$
begin{aligned}
E &= mgh \
implies h &= frac{E}{mg} \
&= frac{1.14912 times 10^{5}~mathrm{J}}{left(65.0~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right)} \
&= 180.21171~mathrm{m} \
h &= boxed{180.~mathrm{m}}
end{aligned}
$$
Notice that since there is no friction, the total mechanical energy is conserved. Also, in both cases, $m$ and $h$ are the same. For the first run, we have
$$
begin{aligned}
KE_text{i,1} + PE_text{i,1} &= KE_text{f,1} + PE_text{f,1} \
0 + mgh &= frac{1}{2}mv_text{f,1}^{2} + 0 \
implies mgh &= frac{1}{2}mv_text{f,1}^{2}
end{aligned}
$$
$$
begin{aligned}
KE_text{i,2} + PE_text{i,2} &= KE_text{f,2} + PE_text{f,2} \
frac{1}{2}mv_text{i,2}^{2} + mgh &= frac{1}{2}mv_text{f,2}^{2} + 0 \
frac{1}{2}mv_text{i,2}^{2} + frac{1}{2}mv_text{f,1}^{2} &= frac{1}{2}mv_text{f,2}^{2} \
implies v_text{f,2}^{2} &= v_text{i,2}^{2} + v_text{f,1}^{2}
end{aligned}
$$
$$
begin{aligned}
v_text{f,2} &< v_text{i,2} + v_text{f,1} \
&< 1.50~mathrm{m/s} + 7.50~mathrm{m/s} \
v_text{f,2} &< 9.00~mathrm{m/s}
end{aligned}
$$
We see that the speed is **less than** $9.00~mathrm{m/s}$.
From the equation we derived, we have
$$
begin{aligned}
v_text{f,2}^{2} &= v_text{i,2}^{2} + v_text{f,1}^{2} \
implies v_text{f,2} &= sqrt{v_text{i,2}^{2} + v_text{f,1}^{2} }\
&= sqrt{left(1.50~mathrm{m/s}right)^{2} + left(7.50~mathrm{m/s}right)^{2}} \
&= 7.64853~mathrm{m/s} \
v_text{f,2} &= boxed{7.65~mathrm{m/s}}
end{aligned}
$$
$$
begin{aligned}
Delta E &= E_text{f} – E_text{i} = E_text{f} – 0 \
&= KE_text{f} + PE_text{f} \
&= frac{1}{2}mv^{2} + mgh \
&= frac{1}{2}left(1865~mathrm{kg}right) left(96.5~mathrm{m/s}right)^{2} + left(1865~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(2420~mathrm{m}right) \
&= 5.29591 times 10^{7}~mathrm{J} \
Delta E &= boxed{5.30 times 10^{7}~mathrm{J}}
end{aligned}
$$
Delta E = 5.30 times 10^{7}~mathrm{J}
$$
$$
begin{aligned}
h_{1} &= v_{y}t + frac{1}{2}gt^{2} = 0 + frac{1}{2}gt^{2} \
h_{1} &= frac{1}{2}gt^{2} \
t^{2} &= frac{2h_{1}}{g} \
implies t &= sqrt{frac{2h_1}{g}}
end{aligned}
$$
$$
begin{aligned}
v_text{f} &= frac{d}{t} \
v_text{f} &= d sqrt{frac{g}{2h_{1}}}
end{aligned}
$$
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
0 + mgh &= frac{1}{2}mv_text{f}^{2} + 0 \
implies h &= frac{1}{2g}v_text{f}^{2} \
&= frac{1}{2g} left(dsqrt{frac{g}{2h_{1}}}right)^{2} \
&= frac{1}{2g} frac{d^{2}g}{2h_{1}} \
&= frac{d^{2}}{4h_{1}} \
&= frac{left(2.50~mathrm{m}right)^{2}}{4left(1.50~mathrm{m}right)} \
&= 1.04167~mathrm{m} \
h &= boxed{1.04~mathrm{m}}
end{aligned}
$$
h = 1.04~mathrm{m}
$$
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
frac{1}{2}mv^{2} + 0 &= 0 + mgh \
v^{2} &= 2gh \
implies v &= sqrt{2gh} \
&= sqrt{2left(9.81~mathrm{m/s^{2}}right) left(2.64~mathrm{m}right)} \
&= 7.19700~mathrm{m/s} \
v &= boxed{7.20~mathrm{m/s}}
end{aligned}
$$
v = 7.20~mathrm{m/s}
$$
$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
0 + mgh_{1} &= frac{1}{2}mv^{2} + mgh_{2} \
v^{2} &= 2gleft(h_{1} – h_{2}right) \
implies v &= sqrt{2gleft(h_{1} – h_{2}right)}
end{aligned}
$$
$$
begin{aligned}
h_{2} &= frac{1}{2}gt^{2} \
t^{2} &= frac{2h_{2}}{g} \
implies t &= sqrt{frac{2h_{2}}{g}}
end{aligned}
$$
$$
begin{aligned}
d &= vt \
&= left(sqrt{2gleft(h_{1} – h_{2}right)} right) left(sqrt{frac{2h_{2}}{g}}right) \
&= sqrt{4h_{2} left(h_{1} – h_{2}right)} \
&= sqrt{4left(0.25~mathrm{m}right) left(1.5~mathrm{m} – 0.25~mathrm{m}right)} \
&= 1.11803~mathrm{m}\
d &= boxed{1.1~mathrm{m}}
end{aligned}
$$
d = 1.1~mathrm{m}
$$
|Organ/Activity | Power in Watts|
|–|–|
| Human metabolism | 80 |
| Human brain | 20 |
| Human heart | 1.33 |
| Biking | 100 |
| Exercise | 50-100 |
| Professional swimming (bursts) | 1200 |
$$
boxed{ 7.7 – 15~mathrm{m/s} }
$$
$$
boxed{2.5 – 25~mathrm{m/s}}
$$
$$
begin{aligned}
t &= 1.0~mathrm{min} times frac{60~mathrm{s}}{1~mathrm{min}} \
t &= 60.~mathrm{s}
end{aligned}
$$
$$
begin{aligned}
E &= Pt \
&= left(8.1~mathrm{W}right) left(60.~mathrm{s}right) \
&= 486~mathrm{J} \
E &= boxed{490~mathrm{J}}
end{aligned}
$$
The correct choice is **C.**
