Energy

Example 1: pulling a wagon of, what else: apples.

The Questions:

a) How much work is done by a force of 20 N on a wagon full of apples with a combined mass of 10 kg if it’s pulled along a frictionless floor for a distance of 10 m?

b) What is the kinetic energy gained by the wagon and apples?

c) If it starts from rest, what is its speed after that distance?

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The Answers:

a) Work is the product of the force and the distance, so

\[\begin{split} \begin{align} W &= F \Delta x \\ W &= (20 \text{ N})(10 \text{ m}) \\ W &= 200 \text{ J} \end{align} \end{split}\]

b) The kinetic energy change is equal to the work (remember, it started from rest):

\[\begin{split} \begin{align} \Delta K &= K-K_0 = W = F\Delta x \\ \Delta K &= K-0 = W = 200 \text{ J} \end{align} \end{split}\]

c) We know the kinetic energy, so we can calculate the velocity:

\[\begin{split} \begin{align} K &= \dfrac{1}{2}mv^2 \\ v &=\sqrt{\frac{2K}{m}} \\ v &= \sqrt{\frac{(2)(200)}{10}} = \sqrt{40} = 6.3 \text{ m/s} \end{align} \end{split}\]

Is this reasonable? That’s about 14 mph. That force of 20 N is not quite 5 pounds. Imagine that you hung a 5 pound weight over a pulley and set the wagon on a sheet of ice (no friction). Here’s the situation drawn about to scale.

I think I could imagine that this thing would be moving pretty fast over that distance.