Problem :

A 10 kg object experiences a horizontal force which causes it to accelerate at 5 m/s² , moving it a distance of 20 m, horizontally. How much work is done by the force?

Problem :

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?

Solution for Problem 2 >>

Recall from our study of uniform circular motion that centripetal force is always directed radially, or toward the center of the circle. Also, of course, the displacement at any given time is always tangential, or directed tangent to the circle:

Work in Uniform Circular Motion

Clearly the force and the displacement will be perpendicular at all times. Thus the cosine of the angle between them is 0. Since W = Fx cosθ , no work is done on the ball.

Close

Problem :

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?

Problem :

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?

Problem :

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?

Solution for Problem 5 >>

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

Because of its weight, mg, the block experiences a force down the incline of magnitude 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.

Close