Q1. A crate is being pulled across a rough floor from rest by force = 36i, 17j w

Question

Q1. A crate is being pulled across a rough floor from rest by force = 36i, 17j when the x-axis is the floor. The crate has a mass M=4.6 Kg and the coefficient of friction with the floor is = 0.61

a- What is the work done by friction?

b- If the crate is pulled a distance, L=4.7m, what is the work done by the force F

c-What is the total work done?

d-What is the final speed of the crate after it has been pulled the distance L?

e-If the time it takes to pull the crate the distance L is t=1.51 seconds, what is the average power dissipated by the frictional force?

can you please solve these questions?

Explanation / Answer

As there is a vertical component of the applied force which will decrease normal reaction.

N = Mg - vertical component of the force applied.

Hence here N = 28.126 N( g = 9.81m/s2).

Hence friction F= uN = 17.156 N

As work done W = F.S = - 17.156L -ve because friction force and displacement are in opposite directions.

Putting L = 4.7, W by friction = - 80.63 Joules

B). Work done by F = FLcos$ where $ is angle between F vector and L vector.

36×4.7 = 169.2 J.

Or you can say that perpendicular component will not do any work.

C). Total work done = algebraic sum of both works = 88.57 J

D). Final speed can be calculated by energy conservation work done = change in kinetic energy hence W = 1/2 mv2

Hence v = 6.205 m/s.

E). Average power = total work done by force / total time

Hence power by friction = - 80.63/1.51 = - 53.39 W.

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