Question

A 55-kg box is being pushed a distance of 7.0 m across the floor by a force whose magnitude is 160 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

Answer 1

Answer:

work done by applied force = 1120J

work done by frictional force = 943.25J

work done by the gravitational force = work done by the normal force = 0J

Explanation:

Given:

Mass of the box = 55kg

Displacement of the box = 7.0m across the floor

Applied force = 160N

Coefficient of kinetic friction ($\mu_k$) = 0.25

Now the work done (W) is given as:

W = F×d

Where,

F = Applied force on the body

d = Displacement of the body in the direction of the applied force

Forces acting on the box are:

1) Applied force

2) Frictional Force

3) Gravitational force

4) Normal force due to the weight

Now the work done by the respective forces:

1) By the applied force

W = 160 kg × 7 m = 1120 J

2) By the frictional force

$W_f =-\mu mg\times displacement$

Where

g = acceleration due to the gravity

and the negative sign here depicts that the frictional force is acting the against the applied force and the direction of displacement.

thus,

$W_f =-0.25\times 55\times 9.8\times 7$

or

$W_f =-943.25J$

The work done by the normal force and the gravitational force will be zero as there is no displacement in the direction of the application of both the forces