Question

A worker pushes a box with a horizontal force of 40.0 N over a level distance of 4.0 m. If a frictional force of 27 N acts on the box in a direction opposite to that of the worker, what net work is done on the box

Answer 1

The net work done on the box is 52.0 J

Step-by-step explanation:

The net work done on the box can be found by subtracting the work done by the frictional force from the work done by the worker. The work done by a force is given by the equation W = F * d * cos(theta), where W is the work done, F is the magnitude of the force, d is the distance, and theta is the angle between the direction of the force and the displacement. In this case, the worker applies a force of 40.0 N and pushes the box over a distance of 4.0 m.

The work done by the worker can be calculated as follows:

W(Worker) = 40.0 N * 4.0 m * cos(0°) = 160.0 J

The work done by the frictional force can be calculated as:

W(Friction) = -27 N * 4.0 m * cos(180°) = -108.0 J

The net work done on the box is the sum of the work done by the worker and the work done by the frictional force:

Net Work = W(Worker) + W(Friction) = 160.0 J - 108.0 J = 52.0 J

Answer 2

Work = (force) x (distance)

The worker does (40N) x (4m) = 160 joules of work.

Friction eats up (27N) x (4m) = 108 joules of that energy,
generating 108 joules of heat.

The remaining (160J - 108J) = 52 joules of energy moves the box.