Work Done by Force: A Physics Problem Solution

[heatherh1545]

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

 

[arunbg]

Hint : What are the different forces that the applied force has to overcome and what are their magnitudes and directions ?
W = F.d

PS: You might also want to delete your double post .

 

[kyle8921]

I would approach this problem by breaking it down into smaller components and using the principles of physics to solve it. Let's start with the first part, finding the work done by the force F.

To find the work done by a force, we use the equation W = F*d, where W is the work, F is the force, and d is the distance moved in the direction of the force. In this case, the force F is applied at an angle of 63.7 degrees with the horizontal, so we need to find the component of the force in the direction of motion, which is F*cos(63.7).

Next, we need to find the distance moved in the direction of the force. Since the block is pushed 1.52 m up the wall, the distance moved in the direction of the force is also 1.52 m. Therefore, the work done by the force F is:

W = F*cos(63.7)*1.52 = 2.8 kg * 9.8 m/s^2 * cos(63.7)*1.52 = 25.34 J

For the second part, finding the work done by the force of gravity, we can use the same equation W = F*d, but this time the force F is the weight of the block, which is mg. The distance d is the same as before, 1.52 m. So, the work done by the force of gravity is:

W = mg*d = 2.8 kg * 9.8 m/s^2 * 1.52 m = 41.62 J

Moving on to the third part, finding the work done by the normal force between the block and the wall. Since the block is pushed up the wall, the normal force is perpendicular to the direction of motion and therefore does no work. So, the work done by the normal force is zero.

Finally, for the fourth part, finding the change in gravitational potential energy, we use the equation ΔU = mgh, where ΔU is the change in potential energy, m is the mass, g is the acceleration due to gravity, and h is the change in height. In this case, the change in height is 1.52 m, so the change in gravitational potential energy is:

ΔU = mgh = 2.8 kg * 9.8 m/s^2 *