A 50-kg box is being pushed a distance of 6.4 m across the floor by a force P with arrow whose magnitude is 154 N. The force P with arrow is parallel to the displacement of the box. The coefficient of kinetic friction is 0.24. 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. (The floor is level and horizontal. Forces that act on the box: pushing force P, frictional force f, normal force N, and weight mg.)

work done by push = 154 * 6.4 J positive

work done by friction = -50*9.81*.24*6.4 J negative

work done by normal = work done by weight = ZERO because force perpendicular to displacement

Well, let's break it down! We have four forces acting on the box: the pushing force P, the frictional force f, the normal force N, and the weight mg.

- The work done by the pushing force P is calculated by multiplying its magnitude by the displacement, so that's 154 N times 6.4 m. The work done by P is 985.6 J.

- The work done by the frictional force f is equal to the product of its magnitude and the displacement. Since the box is moving horizontally, the frictional force is opposing the displacement. Therefore, the work done by f is -(frictional force magnitude times displacement). The frictional force magnitude can be found using the formula f = μN, where μ is the coefficient of kinetic friction and N is the normal force. In this case, N is equal to the weight mg, which is (50 kg times 9.8 m/s^2) N. Therefore, N equals 490 N. Now we can calculate the frictional force: f = (0.24)(490 N). Using this value, we can find the work done by f: -(frictional force magnitude times displacement). This results in -748.8 J.

- The work done by the normal force N is zero because the displacement is perpendicular to the normal force. No work is done.

- Finally, the work done by the weight mg is also zero. The weight acts vertically, which means it's perpendicular to the displacement. No work is done.

So, to sum it up:
- The work done by the pushing force P is 985.6 J.
- The work done by the frictional force f is -748.8 J.
- The work done by the normal force N is 0 J.
- The work done by the weight mg is 0 J.

I hope that didn't "weight" you down with too much information!

To determine the work done on the box by each force, we need to calculate the work done by the pushing force P, the frictional force f, the normal force N, and the weight mg.

1. Work done by the pushing force P:
Since the pushing force is parallel to the displacement of the box, the work done by this force can be calculated using the formula:

Work (W) = Force (F) x Displacement (d) x cos(theta)

Here, the force is given as 154 N, and the displacement is 6.4 m. The angle between the force and displacement is 0 degrees (cos(0) = 1). Therefore, the work done by the pushing force is:

W(P) = 154 N x 6.4 m x cos(0)
W(P) = 985.6 N∙m

2. Work done by the frictional force f:
The frictional force opposes the motion and acts in the opposite direction of the displacement. The work done by friction can be calculated using the formula:

Work (W) = Force (F) x Displacement (d) x cos(theta)

Since the frictional force is given by the equation f = coefficient of kinetic friction x normal force, we need to calculate the normal force first. The normal force N is equal in magnitude and opposite in direction to the weight mg. Therefore, N = mg = 50 kg x 9.8 m/s^2 = 490 N.

Next, the frictional force is given by f = 0.24 x N = 0.24 x 490 N = 117.6 N.

The displacement is 6.4 m, and the angle between the frictional force and the displacement is 180 degrees (cos(180) = -1).

Therefore, the work done by the frictional force is:

W(f) = -117.6 N x 6.4 m x cos(180)
W(f) = -753.6 N∙m

Note: The negative sign indicates that the frictional force is acting in the opposite direction of the displacement.

3. Work done by the normal force N:
The normal force N acts perpendicular to the displacement, so the angle between them is 90 degrees (cos(90) = 0). According to the formula for work, the work done by the normal force is:

W(N) = 0 N x 6.4 m x cos(90)
W(N) = 0 N∙m

Since the normal force does not contribute to the movement of the box, no work is done by the normal force.

4. Work done by weight mg:
The weight mg acts vertically downwards, perpendicular to the displacement. Therefore, the angle between them is 90 degrees (cos(90) = 0). According to the formula for work, the work done by the weight is:

W(weight) = mg x d x cos(90)
W(weight) = 50 kg x 9.8 m/s^2 x 6.4 m x cos(90)
W(weight) = 0 N∙m

Since the weight also does not contribute to the horizontal displacement of the box, no work is done by the weight.

In summary, the work done on the box by each force is as follows:
- The pushing force does +985.6 N∙m of work.
- The frictional force does -753.6 N∙m of work.
- The normal force does 0 N∙m of work.
- The weight does 0 N∙m of work.

To determine the work done by each force, we need to calculate the force applied by each force and multiply it by the distance over which the force is applied.

Force P is parallel to the displacement of the box, which means it does positive work. Its magnitude is given as 154 N, and the distance over which it is applied is 6.4 m. Therefore, the work done by force P is given by:

Work(P) = Force(P) x Distance = 154 N x 6.4 m = 985.6 J

The frictional force, f, is opposing the motion of the box and acts in the opposite direction of the displacement. The magnitude of the frictional force can be calculated using the coefficient of kinetic friction and the normal force, N, which can be determined by the weight of the box.

The weight of the box, mg, is given by:

Weight = mass x acceleration due to gravity = 50 kg x 9.8 m/s^2 = 490 N

The normal force, N, is equal to the weight of the box, so N = 490 N.

The magnitude of the frictional force can be found using the equation:

f = coefficient of kinetic friction x N

f = 0.24 x 490 N = 117.6 N

Since the frictional force is opposite to the displacement, it does negative work. Therefore, the work done by the frictional force is:

Work(f) = -Force(f) x Distance = -117.6 N x 6.4 m = -752.64 J

The normal force, N, acts perpendicular to the displacement, so it does no work. Therefore, the work done by the normal force is zero.

Work(N) = 0 J

The weight, mg, acts vertically downwards and is perpendicular to the displacement, so it does no work as well.

Work(weight) = 0 J

Therefore, the work done by each of the four forces are:

Work(P) = 985.6 J (positive work)

Work(f) = -752.64 J (negative work)

Work(N) = 0 J

Work(weight) = 0 J