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

Answer 2

The work done by both the gravitational force and normal force is equal to zero (0).

Given the following data:

• Mass = 55 kg.

• Distance = 7 meters.

• Magnitude of force = 160 Newton.

• Coefficient of kinetic friction = 0.25.

The forces acting on the box.

In this scenario, there are four forces acting on the box and these include:

• An applied force

• A frictional Force

• A gravitational force

• A normal force due to weight.

For the applied force:

Mathematically, the work done due to an applied force is given by this formula:

$Work\;done = force \times distance\\\\Work\;done = 160 \times 7$

Work done = 1,120 Nm.

For the frictional force:

Mathematically, the work done due to a frictional force is given by this formula:

$Work\;done = -\mu mg \times distance\\\\Work\;done = -0.25 \times 55 \times 9.8 \times 7$

Work done = -943.25.

However, the work done by both the gravitational force and normal force is equal to zero (0) because there isn't any displacement or distance covered in the direction of the application of these forces.

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