Problem related to work Confused between two answers.

[mike-d]

Homework Statement

How much work must you perform to lift a 5 kg bag from the floor to the countertop, 1 meter above the floor?

2. The attempt at a solution

A. You could say that any force greater than the weight of the bag would lift it up. Thus, the weight is 50 N and the force should at least 50 N and above.

since W= F * d * cos(theta)
W = 50 * 1 * cos0 = 50 Joules

B. My classmates insist that this is the correct way. The net work is equal to the change in potential energy, and the total work includes: Work done by weight, and work done by the force...

Work of weight = 50 * 1 * cos180 = -50 Joules
deltaPotential energy = PE2 -PE1 = 50 - 0 = 50 Joules

W(total) = W(weight) + W(force) = deltaPE
-50 + W(force) = 50 joules ==> W(force) = 100 Joules


Can anyone tell me which answer is the correct one? It seems to me that the second answer would be correct if you pulled the bag up suddenly and let it go and it would then stop mid-air at 1 meter. Can anyone verify this? Thanks in advance :)

 

[PhanthomJay]

Your answer is correct. If you are looking for total or net work, that's the change in KE (which is zero). But if you are looking for work done by you, which the question asks, that's the change in PE plus the change in KE which is 50 +0 = +50 J. If you are looking for the work done by the weight, that is the negative of the potential energy change, or -50 J.

 

[mike-d]

Your answer is correct. If you are looking for total or net work, that's the change in KE (which is zero). But if you are looking for work done by you, which the question asks, that's the change in PE plus the change in KE which is 50 +0 = +50 J. If you are looking for the work done by the weight, that is the negative of the potential energy change, or -50 J.

Thanks for the reply :D

However I am confused as to why total work equals to the change in KE...

I would really appreciate it if you could explain it.

 

[tal444]

This involves a lot of formula manipulation...
We know that W = Fd and F = ma.
∴ W = mad

Solving for a from v₂² = v₁² + 2ad gives us
a = (v₂² - v₁²)/2d

plugging that back into W = mad gives us
W = m[(v₂² - v₁²)/2d]d
= \frac{1}{2}m(v₂² - v₁²)
= \frac{1}{2}mv₂² - \frac{1}{2}mv₁²

Notice anything familiar? \frac{1}{2}mv² is our generic equation for KE.
\frac{1}{2}mv₂² - \frac{1}{2}mv₁² gives us ΔKE.

∴ W = ΔKE