Question

A construction worker pushes on a 10 kg crate down an inclined plane that is 7.0 m long and is inclined at an angle of 20 degrees to the horizontal. As a result, the crate slides down the entire length of the plane with a constant speed of 3.0 m/s. Suppose that the worker pushes directly on the crate in a direction parallel to the incline with a force of 50 N for the entire length of the plane. A) What is the work done by the external force of the construction worker? B) What is the work done by kinetic friction? C) What is the coefficient of kinetic friction?

Answer #1

Part A)

External force = 50 N in the direction of motion , SO workdone by it = F * d= 50 * 7 = 350 J

**Workdone =350 J**

Part B)

Workdone by kinetic friction:

First if object is coming with velocity 3 m/s , So kinetic
energy = 1/2 mv^{2} = 35 J .[v is due to friction]

Energy without friction = F * d = mgsin * 7 = 10 *9.8 * sin 20 * 7 = 234.619 .

SO energy Loss due to friction = 234.619 - 35 j=199.6 19 J

**So Workdone by Kinetic frictin = 199.6 19 J**

**Part C)**

So Friction force = 199.617 = 28.517 N

As we know that Friction Force = R = mg cos

So cofficient of kinetic friction = F/mgcos
= 28.517 /(10 *
9.8 * cos 20) = **0.30**

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