How to calculate work done by forces on a block moving up a wall?

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To calculate the work done on a 5kg block moving up a wall, the applied force F must be determined, which is found to be 208 N. The work done by this force is calculated using the formula W=Fcos(theta)(displacement), where the angle used should be 60 degrees, not 30 degrees, since it’s the angle between the force and the direction of motion. The work done by gravity is negative due to the opposing direction of the gravitational force, calculated as W=(9.8 m/s²)cos(180)(3m). The normal force is considered to be zero as it acts perpendicular to the displacement of the block. The increase in gravitational potential energy can be determined using the standard potential energy formula, independent of the angle.
AznBoi
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Okay I kind of get this problem but not completey. Please see if I'm going in the right direction and guide me through it. Thanks!

Here is the Problem: A 5kg block is pushed 3m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of theta=30 degrees with the horizontal. If the coefficient of kinetic friction between block and wall is 0.30, determine the work done by a) F, b) the force of gravity, and c) the normal force between block and wall. d) By how much does the gravitational potential energy increase during this motion?


Here is my work:
First of all I solved for the F force and I got 208 N
So then do I have to plug the force into the equation: W=Fcos(theta)(displacement) ??

Then it would be: 208cos30(3m) right??

For b) you need to find the work force of gravity.
So, W=(9.8m/s^2)cos180(3m)??

for c) normal force between block and wall.
would it be zero?? because the block moves upward and the normal force would be perpendicular to the displacement.

for d) How do you find how much the gravitational potential energy increases by? Do you use Usubi+Ksubi=Usubf+Usuf
 
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You have a problem with the angle. Potential energy increase is the same as always. Your equation is fine.
 
What do you mean I have a problem with the angle?? Are all of the problems that I attempted right?
 
I think he is talking about the work done by the force. 30 degrees is not the angle you should use, if I am understanding this problem correctly.

It is only the component of the force in the direction of the motion that is doing any work.

Dorothy
 
I think 30 degrees is the angle that you need to use for the Fore F because it is being applied diagonally 30 degrees from the horizontal upwards.
 
Just a suggestion, but a closer look at the definition of the angle in that particular formula would be a good idea. Hint: It's the angle between two vectors.

Dorothy
 
oh so if the angle is 30 degrees from the horizontal. and the displacement is 90 degrees. would it be 90-30=60 degrees?
 
That's right.
 
would this be right??

For b) you need to find the work force of gravity.
So, W=(9.8m/s^2)cos180(3m)??

because the force of gravity acts downards and the displacement is upwards so it would be cos180 right?
 
  • #10
Yes, that's right.
 
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