From the definition of the kinetic energy, we have

$$
begin{align*}
KE &= frac{1}{2}mv^{2} \
implies KE &propto v^{2}
end{align*}
$$

The kinetic energy is proportional to the square of the speed.

When the speed is increased by a factor of $2$, the kinetic energy must be $textbf{increased by a factor of $4$.}$

When the speed is increased by a factor of $3$, the kinetic energy must be $textbf{increased by a factor of $9$.}$

In this problem, we explain what happens to the kinetic energy of an object if the work done is positive, such that $W > 0$.

Using work-energy theorem, we have

$$
begin{align*}
Delta KE &= W \
Delta KE &> 0
end{align*}
$$

The change in kinetic energy is positive, so the final kinetic energy must be greater than the initial kinetic energy. The kinetic energy $textbf{increases}$.

In this problem, a pitcher throws a baseball with speed $v = 40~mathrm{m/s}$ and the catcher stops it in her glove. We find the sign of the work done on the ball.

The initial kinetic energy is positive. Since the ball stops, its final kinetic energy is zero. The final kinetic energy is less than the initial. The change in kinetic energy is negative
$$
Delta KE < 0
$$

Using work-energy theorem, we have

$$
begin{align*}
W &= Delta K \
W &< 0
end{align*}
$$

The work done on the ball by the catcher is $textbf{negative}$.

In this problem, we compare the change in potential energy when a box is lifted by $h = 1~mathrm{m}$ off the surface of the Earth and the Moon.

The mass of the object and the change in elevation is the same for both cases. The change in potential energy is

$$
begin{align*}
Delta PE_text{gravity} &= mg Delta h \
Delta PE_text{gravity} &= left( m Delta h right) g \
Delta PE_text{gravity} &propto g
end{align*}
$$

The change in gravitational potential energy is proportional to the acceleration due to gravity in the surface.

Since the acceleration due to gravity on the Earth is greater than the acceleration due to gravity on the Moon, the change in potential energy is $textbf{not the same}$.

From the definition of spring potential energy, we have
$$
begin{aligned}
PE_text{spring} &= frac{1}{2}kx^{2} \
implies PE_text{spring} &propto x^{2}
end{aligned}
$$
The spring potential energy is proportional to the square of the amount of stretching.

When the amount of stretching is increased by a factor of $2$, the potential energy must **increase by a factor of** $4$.

In this problem, a bullet has mass $m = 9.50 times 10^{-3}~mathrm{kg}$ and speed $v = 1.30 times 10^{3}~mathrm{m/s}$. We calculate the kinetic energy.

The kinetic energy, as defined, is

$$
begin{align*}
KE &= frac{1}{2}mv^{2} \
&= frac{1}{2} left( 9.50 times 10^{-3}~mathrm{kg} right) left( 1.30 times 10^{3}~mathrm{m/s} right)^{2} \
&= 8027.50000~mathrm{J} \
KE &= boxed{ 8.03 times 10^{3}~mathrm{J} }
end{align*}
$$

$$
KE = 8.03 times 10^{3}~mathrm{J}
$$

In this problem, a volleyball of mass $m = 0.27~mathrm{kg}$ has kinetic energy $KE = 7.8~mathrm{J}$. We calculate its speed.

From the definition of the kinetic energy, we have

$$
begin{align*}
KE &= frac{1}{2}mv^{2} \
v^{2} &= frac{2KE}{m} \
implies v &= sqrt{frac{2KE}{m}} \
&= sqrt{frac{2 left( 7.8~mathrm{J} right)}{0.27~mathrm{kg}}} \
&= 7.60117~mathrm{m/s} \
v &= boxed{ 7.6~mathrm{m/s} }
end{align*}
$$

$$
v = 7.6~mathrm{m/s}
$$

In this problem, a runner of mass $m = 73~mathrm{kg}$ accelerations from rest $v_text{i} = 0$ to a speed of $v_text{f} = 7.5~mathrm{m/s}$. We calculate the work required for this activity.

First, we find the initial and final kinetic energy of the runner. We have

$$
begin{align*}
KE_text{i} &= frac{1}{2}mv_text{i}^{2} \
KE_text{f} &= frac{1}{2}mv_text{f}^{2} \
end{align*}
$$

We now use the work-energy theorem. We have

$$
begin{align*}
W &= Delta KE = KE_text{f} – KE_text{i} \
&= frac{1}{2}mv_text{f}^{2} – frac{1}{2}mv_text{i}^{2} \
&= frac{1}{2}m left[v_text{f}^{2} – v_text{i}^{2} right] \
&= frac{1}{2} left( 73~mathrm{kg} right) left[ left( 7.5~mathrm{m/s} right)^{2} – 0 right] \
&= 2053.125~mathrm{J} \
W &= boxed{ 2100~mathrm{J} }
end{align*}
$$

$$
W = 2100~mathrm{J}
$$

In this problem, a pinecone of mass $m = 0.14~mathrm{kg}$ is at height $h = 16~mathrm{m}$ above the ground. We calculate the gravitational potential energy. We use $g = 9.81~mathrm{m/s^{2}}$.

From the definition, we have
$$begin{aligned}
PE_text{gravity} &= mgh \
&= left( 0.14~mathrm{kg} right)left( 9.81~mathrm{m/s^{2}} right) left( 16~mathrm{m} right) \
&= 21.9744~mathrm{J} \
PE_text{gravity} &= boxed{ 22~mathrm{J} }
end{aligned}$$

$$PE_text{gravity} = 22~mathrm{J} $$

In this problem, it takes $F = 13~mathrm{N}$ to stretch a certain spring by $x = 9.5 times 10^{-2}~mathrm{m}$. We calculate the potential energy stored in the spring.

First, we calculate the spring constant of the spring. From the definition, we have

$$
begin{aligned}
F &= kx \
implies k &= frac{F}{x} tag{1}
end{aligned}
$$

Substituting this into the equation for potential energy, we have

$$
begin{aligned}
PE &= frac{1}{2}kx^{2} \
&= frac{1}{2} left( frac{F}{x} right)x^{2} \
&= frac{1}{2}Fx \
&= frac{1}{2} left( 13~mathrm{N} right) left( 9.5 times 10^{-2}~mathrm{m} right) \
&= 0.6175~mathrm{J} \
PE &= boxed{ 0.62~mathrm{J} }
end{aligned}
$$

$PE = 0.62~mathrm{J}$