Energy#
Example 2: pushing an already moving wagon of apples#
The Questions:
Back to that wagon of apples. Suppose it was sliding along the ice at a constant speed of 5 m/s and then you started to pull on it with that same force of 20 N from a bit ago.
The Answer:
Here’s how to think about this. The wagon of apples initially had some kinetic energy since it was moving before you pulled on it. The work is equal to the change in kinetic energy. Since there is a force applied, it will now accelerate and after that 10 m will be going faster than the original 5 m/s. How much faster is related to its resulting kinetic energy. So let’s calculate the new kinetic energy after we push:
\[\begin{split}
\begin{align*}
W &= K - K_0 \\
K_0 &= \dfrac{1}{2}mv_0^2 \\
K &= W + 1/2mv_0^2 \\
K &= 200 \text{ J} + \dfrac{1}{2} (10 \text{ kg})(5 \text{ m/s})^2 \\
K &=200 + 125 = 325 \text{ J}
\end{align*}
\end{split}\]
We can now calculate that final speed by taking apart the final \(K\):
\[\begin{split}
\begin{align*}
K &=1/2mv^2 \\
v &=\sqrt{2K/m} \\
v &=\sqrt{(2)(325)/10} \\
v &=\sqrt{65} = 8.1 \text{ m/s}
\end{align*}
\end{split}\]
…which is faster than when we started from rest.