Change in Kinetic Energy for a Sliding Box

AI Thread Summary
A 5.0 kg box slides up a frictionless incline under the influence of a 40 N force, and the problem requires calculating the change in kinetic energy. The initial assumption of zero velocity was incorrect, as it neglected the applied force's impact on kinetic energy. The correct approach involves using the conservation of energy equation, which incorporates both the work done by the applied force and the potential energy change. The final calculation shows that the change in kinetic energy is 232.4 J, confirming the correct application of energy principles. Understanding the role of forces in energy conservation simplifies the problem-solving process.
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Homework Statement


A 5.0 kg box slides up a 10 m long frictionless incline at an angle of 20 degrees with the horizontal, pushed by a 40 N force parallel to the incline. What is the change in kinetic energy?

Homework Equations


Ek = 1/2 mv2
Ep = mgh

The Attempt at a Solution


I tried to assume that vi was zero. Was that incorrect?

1/2 mvf2 + mgh = 0
2.5vf2 = -(5)(9.8)(10/(sin(20)))
vf2= -25.89130...
∴ Ekf = 1/2mvf2 = 64.7 N and ΔEk = 64.7 N

The correct answer is supposed to be 232.4 J. Help?
 
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Well, a couple things. You forgot to involve the 40 N force that is also in the problem -- this will affect the kinetic energy. That's what's messing up your conservation equation and giving you a negative velocity -- it's saying that both kinetic energy and potential energy increase, but that can't be.

To make it easier to solve for the change in kinetic energy, try writing the conservation of energy equation like this: $$\Delta K + \Delta U_{g} = W_{app}$$
This comes from using both energy conservation and the work-kinetic energy theorem.
 
jackarms said:
Well, a couple things. You forgot to involve the 40 N force that is also in the problem -- this will affect the kinetic energy. That's what's messing up your conservation equation and giving you a negative velocity -- it's saying that both kinetic energy and potential energy increase, but that can't be.

To make it easier to solve for the change in kinetic energy, try writing the conservation of energy equation like this: $$\Delta K + \Delta U_{g} = W_{app}$$
This comes from using both energy conservation and the work-kinetic energy theorem.

Sorry, what does ΔUg represent?
 
Oh, that's potential energy from gravity -- mgh. It's just another notation for it. Wapp also means the applied work from the 40 N force, just to clarify that too.
 
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Okay, so then:
If ΔEk + Epf = W
Then:
ΔEk = (40N)(10m) - (5kg)(9.8m/s/s)(10(sin(20)))
ΔEk = 400 - 167.58
ΔEk = 232.410... = 232.4 J

That's the correct answer! Thank you very much. I hadn't thought of just finding the total change in kinetic energy instead of finding them separately.
 
No problem -- glad you could work it out. And don't worry, finding the separate energies is perfectly fine too. Using the change just saves you a few extra steps :)
 
Wouldn't there be a net force up the plane though due to the presence of the 40N force and the horizontal component of the weight in the opposite direction, ((mg)sinθ)?

How come you guys used the work done as Fx instead of (F{net})x?
 
Easy with the question marks. Yes, there are forces involved here, and a net force up the plane, but all the problem asks for is the change in energy, so you can summarize the actions of the forces in energy conservation.
 
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