Please check my solutions -- Mass being pulled with an angled rope

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SUMMARY

The discussion focuses on calculating the work done by a force F of 100 N acting at a 60-degree angle on a block with a mass of 4 kg, while considering a friction coefficient of 0.3. The correct formula for the friction force is identified as f = μN, rather than f = μmg. Additionally, the normal force N must be calculated accurately, taking into account the vertical component of the applied force, which is N - 100sin(60) = 4 * 9.81 N.

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A block of mass M = 4 kg is pulled by a force F = 100 N forming an angle of 60 degrees with themhorizontal plane with friction coefficient 0.3.

Determine the work of force F, friction force and normal force.

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Welcome to PF. :smile:

On your first solution, what are units of "S" for work? What units should Work be in?
 
I assume S to be the displacement?
 
rsk said:
I assume S to be the displacement?
Oh, thanks. I guess the displacement is indeed a variable -- I missed that.
 
The friction force is $$f=\mu N$$ and NOT $$f=\mu mg$$. So to correctly determine the friction force you need to find the normal force N first.
Your equation for the normal force N doesn't seem correct either (though the final result might be correct), it should be $$N-100\sin60=4\cdot 9.81$$ i.e. the force of 100N is actually forming an angle of -60 degrees with the horizontal, that is it is pointing down not up.
 

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