# Is the amount of work done in this case equivalent to both potential and kinetic energy?

This diagram is from a New York State Regent's exam, and it got me thinking about some "obvious" things in energy conservation.

We have someone pulling a rope with 600 N of force at an angle. The problem answer key says that the amount of work done by the person pulling the rope (1800 J) is 330 more than the potential energy added to the box: (9.8 m/s/s)(3m)(50 kg) = 1470 J.

That's all very well, but that got me thinking of what happens to the other 330 J. My first thought was that it goes into kinetic energy for the box. If the box went from rest I am accelerating it at 12 m/s/s against gravity (9.8 m/s/s) for a total acceleration of 2.2 m/s/s, and when I plug that in to figure out how fast the box is moving after the pull, using $$v_f^2 = v_i^2 + 2ax$$ and I get $$v_f^2 = 13.2 \frac{m^2}{s^2}$$, and doing my usual KE equation ($$\frac{1}{2}mv^2$$) and I get 330 J, which is the difference with work done.

So I was just curious if this intuition was correct, because when I looked at the amount of work (1800J) and the amount of PE added to the box (1470 J) and did an inverse sine I got 54.75 degrees as an angle, which made a lot of sense, since my force is at an angle and I could break it into components, and if I do that (take a sine of 54 degrees, multiply by 600N) I get ~490 J and multiplied by the 3m distance I get... 1470 J!

Again, all this makes sense, but I wanted to be sure I was interpreting my mathematical playing around correctly.

• "That's all very well, but that got me thinking of what happens to the other 350 J". What other 350 J are you talking about? Mar 10 at 16:10

The angle you defined (54.75 degrees) is $$\theta = \arcsin(1470\mbox{J}/1800\mbox{J})$$; Then you basically said $$600\mbox{N}\times \sin(\theta)\times 3\mbox{m}=1800\mbox{J}\times\sin(\theta)=1800\mbox{J}\times(1470\mbox{J}/1800\mbox{J})=1470\mbox{J}$$. This is kind of circular calculation, of cause you got back $$1470\mbox{J}$$.

Note that this angle $$\theta$$ is just what you have defined. It has nothing to do with the angle of the force shown in the picture. In the picture, you can arbitrarily change the angle of the force without changing the answer key to the question.

• Thanks, I was thinking what would happen if I change the angle. The force upwards would be the same, so the amount of work done would be the same. But the angle wouldn't make any difference. Mar 10 at 14:21

Again, all this makes sense, but I wanted to be sure I was interpreting my mathematical playing around correctly.

All your "mathematical playing around" up to the last paragraph is correct, if not the most efficient approach. The more direct approach to showing that 330 J is the change in kinetic energy is by using the work energy theorem, discussed below. However, I don't know whether or not that theorem is taught at the high school level in NY (it wasn't many years ago for me when I was in HS in NY).

As @verdelite pointed out, your determination of the angle is based on circular reasoning. The angle of the applied force is completely irrelevant and can be any angle. This can be proven by drawing a free body diagram (FBD) of the box, as shown in the figure below.

Assuming a frictionless pulley, the tension in the rope is the same throughout, namely 600 N, which is applied vertically to the box in the FBD regardless of the angle of the force by the person.

The applicable principle here is the work energy theorem which states:

The net work done on an object equals its change in kinetic energy. Or

$$W_{net}=\Delta KE=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}$$

From the FBD

$$W_{net}=+Td-mgd=+(600)(3)-(490)(3)=+330J$$

Note that work done by gravity is negative since the force of gravity is in the opposite direction of the displacement of the box. The work done by gravity is stored as an increase in the gravitational potential energy of the earth/box system.

Hope this helps.

• Thanks, yes, we do work energy theorem (minus calculus, so we aren't doing force integral but just W = Fd). The thing I realized threw me off was that they had the angle in the problem and I was trying to see if it made any difference to move the person pulling and it dawned on me that especially in the frictionless case it doesn't matter. Basically I was trying to re-derive some things to make sure I set up problems for students correctly. Mar 13 at 17:48
• I should add, the angle bit was something I initially confused with doing certain kinds of force problems, and then I was like, "wait a minute, the tension on the rope will be the same in the vertical direction, D'oh!" Mar 13 at 17:50
• @Jesse Yeah, you might say it's one of those "aha" moments. Mar 13 at 17:54