Mover Pushes 700 kg Piano up 9° Ramp - Force Calculation

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A mover pushes a 700 kg piano up a 9° ramp, reaching a speed of 2 m/s at a height of 3 m. The net force calculated is 73 N, which is the force acting on the piano when it is not accelerating. The confusion arises from the distinction between net force and the force exerted by the mover, which must also overcome gravity. The mover's force is primarily related to the work done against gravity and the kinetic energy gained. The key takeaway is that the question specifically asks for the force when acceleration is zero, which is why the correct answer is 73 N.
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Homework Statement


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

Homework Equations


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##W = \Delta K##

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.
 
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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.
 
paisiello2 said:
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.
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?
 
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
 
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Mr Davis 97 said:
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?
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.
 
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