Question

A 15 kg block is dragged over a rough, hor- izontal surface by a constant force of 107 N acting at an angle of angle 31.9◦ above the horizontal. The block is displaced 59 m and the coefficient of kinetic friction is 0.189.

Find the work done by the 107 N force. The

2

003 (part 2 of 5) 10.0 points

Find the magnitude of the work done by the force of friction.

Answer in units of J.

004 (part 3 of 5) 10.0 points

What is the sign of the work done by the frictional force?

005 (part 4 of 5) 10.0 points

Find the work done by the normal force. Answer in units of J.

006 What is the net work done on the block? Answer in units of J.

Answer 1

Given the mass of the block m = 15kg, the applied force F = 107N, the angle of force applied above the horizontal, the displaceent of the block S = 59m and the coefficient of kinetic friction is = 0.189. The forces acting on the block is shown in the figure below.

1) Only the horizontal component of the applied force can displace the block. So the horizontal component of the applied force is given by,

The work done (W) by the applied force F = 107N is

So the work done by the 107N force is 5359.56J.

2) The frictional force is given by

The magnitude of work done by the force of friction f is

So the magnitude of work done by the frictional force is 1008.90J

3) Since the frictional force is in the opposite direction of displacement, the work done by the frictional force is negative.

4) Since there is no displacement in the y-direction, the work done by the normal force is zero.

5) The net force acting on the block is,

So the work done by the net force is,

The net work done on the block is 4350.66J.