In this problem, we find the case is the most work done by a force $F$ pushed a box a distance $d$ across a level surface at constant speed.

The work done, when the angle between the force and the displacement is $theta$, the work done is

$$
begin{aligned}
W &= Fd cos theta
end{aligned}
$$

The work is maximum when $cos theta$ is maximum, which is when

$$
begin{aligned}
cos theta &= 1 \
implies theta &= cos^{-1} left( 1 right) \
theta &= 0^{circ}
end{aligned}
$$

The work is maximum if $F$ is horizontal, which is option **A.**

### Part B.

For this option, the work in consideration is the work done to raise and lower a set of $m = 25~mathrm{kg}$ barbell to and from the floor ten times. The force exerted is always upward, but the displacement alternates between upward and downward, so the total is **zero** work done.

### Part C.

For this option, the work in consideration is the work done to hold a set of $m = 25~mathrm{kg}$ barbells at constant height for $t = 3~mathrm{min}$. The displacement is zero, so the work done must be **zero**.

### Part D.

For this option, the work in consideration is the work done to kick a set of $m = 25~mathrm{kg}$ barbell and cause them to roll across the floor. The barbell gains kinetic energy from rest, so the work done is **positive.**

For this problem, we are given the energy graph of a brick dropped from a height $h = 8~mathrm{m}$. We approximate the mass of the brick. We use $g = 9.81~mathrm{m/s}$.

The initial potential energy from the graph is $PE = 80~mathrm{J}$. This is the potential energy when the height is $h$.

A. $1~mathrm{kg}$

From the graph, at$quad t=1s$ ,

$PE=(30)J$

$KE=(50)J$

so,$quad F=(PE+KE)=(80)J$

$$
boxed{text{ans :}rightarrow(C)}
$$

From the graph, at$quad t=1squad,quad KE=(50)J=dfrac{1}{2}mv^2$

$Rightarrowquad v=Big(sqrt{dfrac{2times50}{m}}Big)m/s=Big(sqrt{dfrac{100}{1}}Big)m/s=10m/s$

$$
boxed{text{Ans : }rightarrow(B)}
$$

In this problem, we are asked which situation requires the greatest average power. We use $g = 9.81~mathrm{m/s^{2}}$.

### Part A.

For this option, the action is lifting a block of mass $m = 5~mathrm{kg}$ to a height of $h = 2~mathrm{m}$ in $t = 2~mathrm{s}$. The work done must be equal to change in potential energy of the block.

$$
W = mgh
$$

From the definition of power, we have

$$
begin{aligned}
P_{A} &= frac{W}{t} \
&= frac{mgh}{t} \
&= frac{left(5~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(2~mathrm{m}right)}{2~mathrm{s}} \
P_{A }&= 49.05~mathrm{W}
end{aligned}
$$

### Part B.

For this option, a block is pushed across a level surface with a net force of $F = 10~mathrm{N}$ at a velocity of $v = 3~mathrm{m/s}$.

From another definition of power, we have

$$
begin{aligned}
P_{B} &= Fv \
&= left(10~mathrm{N}right) left(3~mathrm{m/s}right) \
P_{B} &= 30~mathrm{W}
end{aligned}
$$

### Part C.

For this option, a rolling wheel changes kinetic energy from $KE_text{i} = 15~mathrm{J}$ to $KE_text{f} = 55~mathrm{J}$ in $t = 20~mathrm{s}$. The work done is equal to the change in kinetic energy, so

$$
W = Delta KE = KE_text{f} – KE_text{i}
$$

The power must be

$$
begin{aligned}
P_{C} &= frac{W}{t} = frac{KE_text{f} – KE_text{i}}{t} \
&= frac{55~mathrm{J} – 15~mathrm{J}}{20~mathrm{s}} \
P_{C} &= 2~mathrm{W}
end{aligned}
$$

### Part D.

For this option, a lightbulb of power $P = 10~mathrm{W}$ for $t = 20~mathrm{h}$. From the given, we already know that

$$
P_{D} = 10~mathrm{W}
$$

Based on the calculatations, option **A** has the greatest average power, with $P = 49.05~mathrm{W}$.

In this problem, we are given that a ball of mass $m = 2.0~mathrm{kg}$ from a height of $h = 4.0~mathrm{m}$ loses $10%$ of its mechanical energy to thermal energy when it hits the floor. We calculate (a) the kinetic energy jsut before it hits the floor, (b) the kinetic energy right after the it hits the floor, and (c) the height to which the ball returns after it returns bounce. We assume no air resistance and use $g = 9.81~mathrm{m/s^{2}}$.

### Part A.

First, we calculate the total mechanical energy $E_{1}$. The ball starts from rest, so this must be equal to the initial gravitational potential energy.

$$
E_{1} = mgh
$$

Just before it hits the floor, all of the mechanical energy is in the form of the kinetic energy, hence

$$
begin{aligned}
KE_text{a} &= mgh \
&= left(2.0~mathrm{kg}right) left(9.81~mathrm{m/s^{2}}right) left(4~mathrm{m}right) \
&= 78.48~mathrm{J} \
KE_text{a} &= boxed{78~mathrm{J}}
end{aligned}
$$

### Part B.

The kinetic energy jut after hitting hte floor is equal to $90%$ of the kinetic energy just before hitting hte floor. Hence,

$$
begin{aligned}
KE_text{b} &= 0.90KE_mathrm{a} \
&= 0.90left(78.48~mathrm{J}right) \
&= 70.632~mathrm{J} \
KE_text{b} &= boxed{71~mathrm{J}}
end{aligned}
$$

### Part C.

The total mechanical energy for this part must be $90%$ of the initial total mechanical energy. The potential energy at the top must be equal to the total mechanical energy.

$$
begin{aligned}
E_{2} &= 0.90E_{1} \
mgh_{2} &= 0.90mgh_{1} \
h_{2} &= 0.90h_{1} \
&= 0.90left(4.0~mathrm{m}right) \
h_{2} &= boxed{3.6~mathrm{m}} \
end{aligned}
$$

$$
begin{aligned}
KE_text{a} &= 78~mathrm{J} \
KE_text{b} &= 71~mathrm{J} \
h_{2} &= 3.6~mathrm{m}
end{aligned}
$$