physics help?

Question

i need help with this i just cant figure it out

a 17.4 kg block is dragged over a rough, horizontal surface by a constant force of 185 N acting at an angle of 32.9 degrees above the horizontal. the block is displaced 58.2 m, and the coefficient of kinetic friction is .17
the acceleration of gravity is 9.8 m/s^2

find the work done by the 185 N force. answer in units of J (joules)

find magnitude of the work done by the force of friction, answer in J

work done by the normal force? answer in J

net work done on block? in J


if you can help me solve it it would help or at lease give me some instruction how to with some equations

Thanks

Answer 1

The key equations are:
Work = Force * distance
Fn = mg sin θ
Ff = μFn

1.) The work done by the 185N force is the easiest part.
Work done = F * d = 185 N * 58.2m = 10767 J

2 and 3 require you to find the normal force (Fn), which is the force pushing the block into the surface.

Fn = mg sin θ
mg = weight of block
sin θ is used to find the portion of the weight that pushes into the surface that is at an angle.

Fn = 17.4 kg * 9.8 m/s^2 * sin 32.9 = 92.622N

Ff = μFn
μ = coefficient of friction

Ff = .17 * 92.622N = 15.746 N
2.) Work done by Ff = Ff * d = 15.746 N * 58.2 m = 916.4 J

3.) Work done by Fn is 0 J, because the force acts perpendicular to the movement of the block - the block does not move in the direction of Fn.

Net work done is found by adding up the work.
Work done by 185N force = 10767 J
Work done by friction = 916.4 J
4.) Total work = W - Wf = 10767 J - 916.4 J = 9850.6 J