In this problem, we calculate the change in kinetic energy of the box in Guided Example 6.7. The work done on the box is $W_text{total} = 19.9~mathrm{J}$.

The change in kinetic energy is $textbf{equal}$ to $19.9~mathrm{J}$. From the work-energy theorem, we have

$$
begin{align*}
Delta KE &= W_text{total} \
Delta KE &= boxed{19.9~mathrm{J}}
end{align*}
$$

In this problem, an object of mass $m = 0.40~mathrm{kg}$ is falling with speed $v_text{i} = 6.0~mathrm{m/s}$ at time $t = 1.0~mathrm{s}$. At $t = 2.0~mathrm{s}$, it has kinetic energy of $KE_text{f} = 25~mathrm{J}$.

Part A.

For this part, we calculate the kinetic energy at time $t = 1.0~mathrm{s}$. The mass is $m$ and the speed is $v_text{i}$.

$$
begin{align*}
KE_text{i} &= frac{1}{2}mv_text{i}^{2} \
&= frac{1}{2} left( 0.40~mathrm{kg} right) left( 6.0~mathrm{m/s} right)^{2} \
KE_text{i} &= boxed{ 7.2~mathrm{J} }
end{align*}
$$

Part B.

For this part, we calculate the speed at $t = 2.0~mathrm{s}$. The mass is $m$ and the kinetic energy is $KE_text{f}$. We have

$$
begin{align*}
KE_text{f} &= frac{1}{2}mv_text{f}^{2} \
v_text{f}^{2} &= frac{2KE_text{f}}{m} \
v_text{f} &= sqrt{frac{2KE_text{f}}{m}} \
&= sqrt{frac{2left( 25~mathrm{J} right)}{0.40~mathrm{kg}}} \
&= 11.18033~mathrm{m/s} \
v_text{f} &= boxed{ 11~mathrm{m/s} }
end{align*}
$$

Part C.

For this part, we calculate the work done on the object between $t = 1.0~mathrm{s}$ and $t = 2.0~mathrm{s}$. From the work-energy theorem, we have

$$
begin{align*}
W &= KE_text{f} – KE_text{i} \
&= 25~mathrm{J} – 7.2~mathrm{J} \
&= 17.8~mathrm{J} \
W &= boxed{ 18~mathrm{J} }
end{align*}
$$

begin{enumerate}
item [(a)] $KE_text{i} = 7.2~mathrm{J}$
item [(b)] $v_text{f} = 11~mathrm{m/s}$
item [(c)] $W = 18~mathrm{J}$
end{enumerate}