# Work done by a mover

1. Apr 22, 2015

### Mr Davis 97

1. The problem statement, all variables and given/known data

A mover pushes a 700 kg piano from rest up a ramp with a 9° slope. The piano is travelling at 2 m/s when it is 3 m above the ground. How much force does the mover exert on the piano?

2. Relevant equations

$W = \Delta K$

3. The attempt at a solution

I started with the equation $F_{net}d\cos\theta = \frac{1}{2}mv^{2}$

Then I solved for the net force: $F_{net} = \displaystyle\frac{mv^{2}}{2d\cos\theta}$

I then solved for displacement: $d = \displaystyle\frac{3}{\sin9^{\circ}} = 19.2$

Next, I found the net force: $F_{net} = 73~N$

This is the net force on the body, so I found the force that the mover exerted against gravity to find what he exerted, which came out to me $1150~N$. However, this is this the incorrect answer. My textbook says that the correct answer is 73 N. But isn't this the net force on the body, and not the force that the mover exerts? Please explain what I am doing wrong.

2. Apr 22, 2015

### paisiello2

Even though gravity also does work, this work gets stored as potential energy in the system. So the work done by gravity simply cancels out with this potential energy.

3. Apr 22, 2015

### Mr Davis 97

I still don't understand why I am wrong. Doesn't the mover have to overcome the parallel force of gravity in addition to applying his own force to accelerate the piano?

4. Apr 22, 2015

### paisiello2

Yes, but this is stored as potential energy in the system.

PEi + KEi +Work done by mover + Work done by gravity = PEf + KEf
0 + 0 + Work done by mover + (mgsinθ)(h/sinθ) = mgh + 1/2mv2
Work done by mover = 1/2mv2

Last edited: Apr 22, 2015
5. Apr 22, 2015

### billy_joule

For the initial acceleration more than 73N will need to be applied but that is not the question. You are only asked about the force needed when acceleration is zero.