Solve the Homework Problem with Work and Energy

AI Thread Summary
The discussion focuses on solving a physics problem involving work and energy related to a crate being pushed up a slope. The user attempts to apply the conservation of energy principle, equating work done by a force to the change in kinetic energy and potential energy. However, there is confusion regarding the correct formulation of gravitational work, with an incorrect expression involving sin(20 degrees) being highlighted. The correct potential energy should be calculated using the height directly, not adjusted by the sine of the angle. Clarification is sought on the validity of both approaches, with emphasis on the proper application of work-energy principles.
Anthonyphy2013
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Homework Statement



Doug pushes a 5.30 kg crate up a 2.20-m-high 20 degrees frictionless slope by pushing it with a constant horizontal force of 27.0 N. What is the speed of the crate as it reaches the top of the slope?

a) Solve this problem using work and energy.



Homework Equations


Wd=delta KE, ke=.5mv^2 and pe=mgh

The Attempt at a Solution


Conservation of energy
W.d=Pe+KE
Fcos20=mg(h/sin20)+.5mv^2
v= 3.9m/s
but I consider another way about the work done
work done = delta KE
work done on force = Fcos20
work done against the gravitational force = mg(h/sin20)
so Work done by force - work done against the gravitational force =Change of KE .
which one is a corrected concept or both make sense. thanks so much.
 
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Anthonyphy2013 said:
...Fcos20=mg(h/sin20)+.5mv^2...

This is not correct.
 
szynkasz said:
This is not correct.

why is that not corrected ?
Since work done must be equal to delta KE , so we consider workd done on gravitational force is not a conserved work done ?
 
Last edited:
Left side of the equation is force, right is energy. Potential energy is ##mgh##, not ##mg\frac{h}{\sin 20^o}##
 

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