Solving Physics: 2.8 kg Pushed 1.52 m Up Wall w/F, G, & N Forces

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SUMMARY

The discussion focuses on calculating the work done by various forces acting on a 2.8 kg block pushed 1.52 m up a vertical wall with a constant force F applied at an angle of 63.7 degrees. Key calculations include the work done by the applied force F, the gravitational force, and the normal force, as well as the increase in gravitational potential energy. The coefficient of kinetic friction is noted as 0.586, which plays a crucial role in determining the frictional force affecting the block's motion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of work-energy principles
  • Familiarity with free body diagrams
  • Basic grasp of friction coefficients and their applications
NEXT STEPS
  • Calculate work done by a force using the formula W = F * d * cos(θ)
  • Explore gravitational potential energy calculations and the formula ΔPE = m * g * h
  • Learn about normal force and its role in frictional force calculations
  • Study the effects of angles on force components in physics problems
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in solving problems related to forces and motion in vertical scenarios.

heatherh1545
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I have spent the last hour on this problem and I just can't seem to come up with the right answer:A 2.8 kg is pushed 1.52 m up a vertical wall with constant force of magnitude F applied at an angle of 63.7 degrees with the horizontal. The acceleration of gravity is 9.8 m/s (seconds squared) If the coefficient of kinetic friction between the block and the was is .586 find the:
a) work done by F
b) Find the work done by the force of gravity
c) Find the work done by the normal force between the block and the wall
d) by how much does the gravitational potential energy increase during this motion?

I assumed that I was on the right track when I was trying to find the force by using : F=mu*mg/ cos theta + mu sin theta
 
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I think you need to revisit your free body diagram...

The force driving the object will break down into two component forces, one in the direction of the motion (the one that does the work) and one perpendicular to the wall (the normal force for the calculation of the frictional force).

By the looks of things, I think you just made an algebra error
 
heatherh1545 said:
I assumed that I was on the right track when I was trying to find the force by using : F=mu*mg/ cos theta + mu sin theta
But you are not asked to find F. You are given F. You are asked to find the work done by F.

As Fred points out, your analysis of the forces is incorrect. What are the horizontal forces? What is the net horizontal force? What is the downward force? What is the upward force? What is the acceleration?

AM
 

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