Work Done by Gravity: Find Magnitude

AI Thread Summary
The discussion revolves around calculating the work done by gravity on a 2.52kg block pushed up a vertical wall. The initial calculation using W = mgh resulted in a negative value, which was identified as incorrect. The correct approach involves using the formula W = Fd cos(theta), where theta is the angle between the force and displacement. Participants emphasized the importance of understanding the relationship between force direction and displacement in work calculations. Clarification on the definition of gravitational potential energy and its implications for work done by gravity was also suggested.
AdnamaLeigh
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I'm sick of this stuff

A 2.52kg block is pushed 1.54m up a vertical wall with constant speed by a constant force of magnitude of F applied at 63.6 degrees with the horizontal. The coefficient of friction is .574. Find the magnitude of the work done by the force of gravity.

This is what I did:
W=mgh
W=2.52(-9.8)(1.52)=-38.032J

Apparently it's wrong and I don't understand why. I drew the diagram and the force of gravity is going down.

I already found F=38.648N and the work done by F, 53.265J for a previous problem but I don't think those results will help.
 
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Please tell me what mgh is equal to, and you will have solved your problem. (I don't mean the number, I mean look it up in your book and see what its called). :-)
 
AdnamaLeigh,
the work, W, done by a force, F, acting on a body while it is displaced through a distance d (a long a straight line), is given by
W = Fd cos(theta)
where theta is the angle between the the force and the displacement. The direction of the force and the displacement does not come into consideration in this calculation of the work.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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