**Problem1 :**

A 10 kg object experiences a horizontal force which causes it to accelerate at5 m/s

^{2}, moving it a distance of 20 m, horizontally. How much work is done by the force?
The magnitude of the force is given by

*F*=*ma*= (10)(5) = 50 N. It acts over a distance of 20 m, in the same direction as the displacement of the object, implying that the total work done by the force is given by*W*=*Fx*= (50)(20) = 1000 Joules.**Problem2 :**

A ball is connected to a rope and swung around in uniform circular motion. The tension in the rope is measured at 10 N and the radius of the circle is 1 m. How much work is done in one revolution around the circle?

Work in Uniform Circular Motion

*W*=

*Fx*cos

*θ*, no work is done on the ball.

**Problem3 :**

A crate is moved across a frictionless floor by a rope THAT is inclined 30 degrees above horizontal. The tension in the rope is 50 N. How much work is done in moving the crate 10 meters?

In this problem a force is exerted which is not parallel to the displacement of the crate. Thus we use the equation

*W*=*Fx*cos*θ*. Thus*W*=

*Fx*cos

*θ*= (50)(10)(cos 30) = 433 J

**Problem4 :**

A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight?

The crate, and thus the point of application of the force, does not move. Thus, though a force is applied, no work is done on the system.

**Problem5 :**

A 5 kg block is moved up a 30 degree incline by a force of 50 N, parallel to the incline. The coefficient of kinetic friction between the block and the incline is .25. How much work is done by the 50 N force in moving the block a distance of 10 meters? What is the total work done on the block over the same distance?

Finding the work done by the 50 N force is quite simple. Since it is applied parallel to the incline, the work done is simply

*W*=*Fx*= (50)(10) = 500 J.
Finding the total work done on the block is more complex. The first step is to find the net force acting upon the block. To do so we draw a free body diagram:

Work on an Incline

*mg*sin 30 = (5)(9.8)(.5) = 24.5 N . In addition, a frictional force is felt opposing the motion, and thus down the incline. Its magnitude is given by

*F*

_{k}=

*μF*

_{N}= (.25)(

*mg*cos 30) = 10.6 N . In addition, the normal force and the component of the gravitational force that is perpendicular to the incline cancel exactly. Thus the net force acting on the block is: 50 N -24.5 N -10.6 N = 14.9 N , directed up the incline. It is this net force that exerts a ìnet workî on the block. Thus the work done on the block is

*W*=

*Fx*= (14.9)(10) = 149 J