How Is Work Calculated When a Block Is Pushed Up a Wall?

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A 2.79 kg block is pushed 1.48 m up a vertical wall at a constant speed with a force applied at a 60.8-degree angle. The work done by the applied force F needs to be calculated while disregarding friction initially. The initial calculation attempted to use gravitational force and the angle but was incorrect as it did not account for the force F properly. The discussion emphasizes the importance of considering all forces, including friction, for accurate calculations. Clarification on the net work done by all forces is also sought, indicating a need for a comprehensive understanding of the problem.
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1. A 2.79 kg block is pushed 1.48 m up a vertical
wall with constant speed by a constant force
of magnitude F applied at an angle of 60.8
with the horizontal.
The acceleration of gravity is 9.8 m/s2 .If the coefficient of kinetic friction between
the block and wall is 0.581, find the work done
by F



2W=FD



3. seems pretty straight foward but i guess I am missing something. first attempt was W= (m)(g)(D)

W= (2.79kg)(9.8m/s^2)(cos(90-60.8))(1.48)
W=35.3J
that answer seems reasonable, but is not correct. first question is only asking what the work done by F is disregarding the friction. next question is about the work done by friction.(
 
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Idividebyzero said:
first question is only asking what the work done by F is disregarding the friction.
I don't understand this comment. In any case, why not start by finding F? (Don't disregard friction!)
 
oh crud. i was getting this problem mixed up with another one no wonder the answer was wrong. it is asking for the net work
 
Last edited:
bump this is the last one giving me problems
 
Idividebyzero said:
it is asking for the net work
The net work done by all forces? No calculation needed for that.

Can you explain the calculation you did in the first post?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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