All Solutions
Page 311: Standardized Test Practice
textbf{Given:} \
$m = 55 text{kg}$ \
$v_i = 4.0 frac{text{m}}{text{s}}$ \
$v_f = 6.0 frac{text{m}}{text{s}}$ \
textbf{Calculation:} \
The work done by the bicyclist can be expressed as the change in its kinetic energy thus we can express it using the equation below
$$
W = Delta KE = KE_f – KE_i
$$
$$
W = frac{1}{2}mv_f^2 – frac{1}{2}mv_i^2
$$
$$
W = frac{1}{2}m(v_f^2 – v_i^2)
$$
Plugging in the given values, we have
$$
W = frac{1}{2} cdot (55) cdot ((6.0)^2 – (4.0)^2)
$$
$$
boxed{W = 550 text{J}}
$$
W = 550 text{J}
$$
$W = Delta K = (1/2) * (55) * (6.0^2 – 4.0^2)$
$$
W = 550 J
$$
550 J
$$
textbf{Given:} \
$m = 4.0 text{kg}$ \
$h = 2.5 text{m}$ \
Since the ball is initially at rest, the mechanical energy $E$ of the system is equal only to its initial potential energy $PE$.
$$
E = PE
$$
$$
E = mgh
$$
We know that the kinetic energy is given by
$$
KE = frac{1}{2}mv^2
$$
Since the maximum kinetic energy of the system is equal to its mechanical energy, we can express it as follows
$$
KE_text{max} = E
$$
$$
frac{1}{2}mv_text{max}^2 = mgh
$$
Isolating $v_text{max}$ on one side of the equation, we have
$$
v = sqrt{2gh}
$$
Plugging in the given values
$$
v = sqrt{2 cdot (9.8) cdot (2.5)}
$$
$$
boxed{v = 7.0 frac{text{m}}{text{s}}}
$$
v = 7.0 frac{text{m}}{text{s}}
$$
$m g h = (1/2) m v^2$
Solve for v:
$$
v = sqrt{2 g h} = sqrt{(2)*(9.80)*(2.5)} = 7.0 m/s
$$
7.0 m/s
$$
textbf{Given:} \
$m = 4.5 text{kg}$ \
$Delta h = 1.5 text{m}$ \
textbf{Calculation:}\
The work done on the box is equal to the change in potential energy of the box after being lifted from the floo. We can express this using the equation below.
$$
W = Delta PE
$$
$$
W = mgDelta h
$$
Plugging in the given values
$$
W = (4.5) cdot (9.8) cdot (1.5)
$$
$$
boxed{W = 66 text{J}}
$$
W = 66 text{J}
$$
$W = (4.5)*(9.80)*(1.5 – 0)$
$$
W = 66 J
$$
66 J
$$
textbf{Given:} \
$m = 6.0 times 10^{-2} text{kg}$ \
$h = 1.0 text{m}$ \
$E_text{loss} = 0.14 text{J}$ \
textbf{Calculation:}\
Since the ball is initially at rest, the mechanical energy $E$ of the system is equal only to the initial potential energy $PE$ of the ball.
$$
E = PE
$$
$$
E = mgh
$$
$$
E = mgh – E_text{loss}
$$
$$
KE = E
$$
$$
KE = mgh – E_text{loss}
$$
Plugging in the given values, we now have
$$
KE = (6.0 times 10^{-2}) cdot (9.8) cdot (1.0) – 0.14
$$
$$
boxed{KE = 0.45 text{J}}
$$
KE = 0.45 text{J}
$$
$K_b = PE = m g (h)$
$K_b = (6.0 times 10^{-2})*(9.80)*(1.0)$
$K_b = 0.588 J$
The ball’s kinetic energy just after it bounces off the surface:
$$
K_a = K_b – 0.14 = 0.588 – 0.14 = 0.45 J
$$
0.45 J
$$
textbf{Given:} \
$m = 2.5 text{kg}$ \
$h_i = 1.2 text{m}$ \
$h_f = 2.6 text{m}$ \
textbf{Calculation:}\
The change in potential energy is given by the equation below
$$
Delta PE = mgDelta h
$$
$$
Delta PE = mg(h_f – h_i)
$$
Plugging in the given values, we have
$$
Delta PE = (2.5)(9.8)(2.6 – 1.2)
$$
$$
boxed{Delta PE = 34 text{J}}
$$
Delta PE = 34 text{J}
$$
$W = (2.5)*(9.80)*(2.6 – 1.2)$
$$
W = 34 J
$$
34 J
$$
textbf{Given:} \
$m = text{mass of the ball}$ \
$v_1 = text{speed of the ball}$ \
$E_2 = frac{1}{2}E_1$ \
textbf{Calculation:}\
The energy of the ball is purely kinetic thus the energy of the ball before and after bouncing off the wall can be written as
begin{equation}
E_1 = frac{1}{2}mv_1^2
end{equation}
begin{equation}
E_2 = frac{1}{2}mv_2^2
end{equation}
Since $E_2 = frac{1}{2}E_1$, we can write Equation (2) as follows
$$
frac{1}{2}mv_2^2 = frac{1}{2} left( frac{1}{2}mv_1^2 right)
$$
Isolating $v_2$ on one side of the equation
$$
v_2 = frac{v_1}{sqrt{2}}
$$
Rationalizing the denominator, we have
$$
v_2 = frac{v_1}{sqrt{2}} cdot frac{sqrt{2}}{sqrt{2}}
$$
$$
boxed{v_2 = frac{sqrt{2}}{2}v_1}
$$
v_2 = frac{sqrt{2}}{2}v_1
$$
$dfrac{v_2}{v_1} = sqrt{dfrac{K_2}{K_1}} = sqrt{dfrac{1}{2}} = dfrac{1}{sqrt{2}} = dfrac{sqrt{2}}{2}$
$$
v_2 = dfrac{sqrt{2}}{2} v_1
$$
dfrac{sqrt{2}}{2} v_1
$$
textbf{Given:} \
$v = 14 frac{text{}}{text{}}$ \
textbf{Calculation:}\
Since the mechanical energy of the ball is conserved, its potential energy at the top of the track must be equal to its velocity before reaching the ground. We can epxress this using equation below
$$
PE_text{max} = KE_text{max}
$$
$$
mgh = frac{1}{2}mv^2
$$
Isolating $h$ on one side of the equation
$$
h = frac{v^2}{2g}
$$
Plugging in the given values, we have
$$
h = frac{(14)^2}{2(9.8)}
$$
$$
boxed{h = 10 text{m}}
$$
h = 10 text{m}
$$
$m g h = (1/2) m v^2$
Cancel m:
$g h = (1/2) v^2$
Solve for h:
$$
h = dfrac{v^2}{ 2 g} = dfrac{(14)^2}{(2)*(9.8)} = 10 m
$$
10 m
$$
textbf{Given:} \
$m = 1.0 text{kg}$ \
$W = 4.9 text{J}$ \
textbf{Calculation:}\
The work done by the spring to change the potential energy of the box is given by the equation below
$$
W = Delta PE
$$
$$
W = mgDelta h
$$
Isolating $Delta h$ on one side of the equation, we have
$$
Delta h = frac{W}{mg}
$$
Plugging in the given values
$$
Delta h = frac{4.9}{(1.0) cdot (9.8)}
$$
$$
boxed{Delta h = 0.5 text{m}}
$$
Delta h = 0.5 text{m}
$$
$P = 4.9 J$
$m g h = 4.9$
$(1.0)*(9.8)*(h) = 4.9$
Thus:
$$
h = dfrac{4.9}{(1.0)*(9.8)} = 0.5 m
$$
0.5 m
$$
