A 50-kg box is being pushed a distance of 7.1 m across the floor by a force whose magnitude is 175 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.19. 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.
Well, it seems like the box is getting quite the workout! Let's break down the forces and calculate the work done by each:
1. The applied force: The magnitude of the force is 175 N, and it is parallel to the displacement of the box. So, the work done by this force is W = F * d = 175 N * 7.1 m = 1232.5 J (positive, because it is doing work on the box).
2. The force of gravity: This force is acting vertically downwards, perpendicular to the displacement of the box. Since the box is not moving vertically, the work done by gravity is zero because there is no displacement in that direction.
3. The normal force: The normal force is equal in magnitude but opposite in direction to the force of gravity. Since the box is not moving vertically, the displacement in the direction of the normal force is zero. Therefore, the work done by the normal force is also zero.
4. The force of friction: The coefficient of kinetic friction is given as 0.19. The formula for the force of friction is F_friction = coefficient_friction * normal_force. Since the normal force is equal to the force of gravity (mg), the force of friction is 0.19 * 50 kg * 9.8 m/s^2 = 94.1 N. The work done by friction is W = F_friction * d = 94.1 N * 7.1 m = 668.11 J (negative, because friction opposes the direction of motion).
So, the work done by each force is as follows:
1. Applied force: +1232.5 J
2. Force of gravity: 0 J
3. Normal force: 0 J
4. Force of friction: -668.11 J
And there you have it! The work done by each of the four forces acting on the box. Just remember, work may make you tired, but solving physics problems can be a real workout for the brain!
To find the work done on the box by each of the four forces, we need to calculate the work done by each force separately.
1. Work done by the applied force:
The work done by the applied force is given by the formula:
Work = Force x Displacement x cos(theta)
where theta is the angle between the force and the displacement. Since the force and displacement are parallel in this case, the angle theta is 0 degrees.
Work = 175 N x 7.1 m x cos(0°)
= 175 N x 7.1 m x 1
= 1232.5 Nm or 1232.5 Joules
So, the work done by the applied force is 1232.5 Joules (positive value because the applied force is doing work on the box).
2. Work done by the force of gravity:
The force of gravity is acting vertically downward on the box. The work done by gravity can be calculated using the formula:
Work = Force x Displacement x cos(theta)
In this case, since the force of gravity is acting perpendicular to the displacement, the angle theta is 90 degrees. The work done by gravity is then:
Work = mg x d x cos(90°)
= 50 kg x 9.8 m/s^2 x 7.1 m x cos(90°)
= 0 Nm or 0 Joules
So, the work done by the force of gravity is 0 Joules (zero value because the force of gravity is perpendicular to the displacement and does not do any work on the box).
3. Work done by the normal force:
The normal force is acting perpendicular to the floor and does not contribute to the displacement of the box. Therefore, the work done by the normal force is also 0 Joules.
4. Work done by the friction force:
The work done by the friction force can be calculated using the formula:
Work = Force x Displacement x cos(theta)
In this case, the friction force is acting opposite to the direction of motion (displacement) of the box, so the angle theta between the force and the displacement is 180 degrees.
Work = friction force x displacement x cos(180°)
= [μ * (normal force)] x d x cos(180°)
= [0.19 x (50 kg x 9.8 m/s^2)] x 7.1 m x cos(180°)
= - 670.79 Nm or -670.79 Joules
Notice the negative sign, which indicates that the work done by the friction force is negative. This means that the work is being done against the motion of the box.
To summarize:
- Work done by the applied force = 1232.5 Joules (positive)
- Work done by the force of gravity = 0 Joules (zero)
- Work done by the normal force = 0 Joules (zero)
- Work done by the friction force = -670.79 Joules (negative)
To determine the work done on the box by each of the four forces, we need to calculate the work done by each force separately.
1. Work done by the applied force:
The applied force is the force that is pushing the box across the floor. The work done by this force can be calculated using the formula:
Work = Force * Distance * Cos(theta)
In this case, the magnitude of the applied force is 175 N, and the distance traveled by the box is 7.1 m. Since the force is parallel to the displacement, the angle (theta) between the force and displacement is 0 degrees. Therefore, the cosine of 0 degrees is 1, and the formula simplifies to:
Work = 175 N * 7.1 m * 1 = 1232.5 J
So, the work done by the applied force is 1232.5 Joules.
2. Work done by the gravitational force:
The gravitational force is acting vertically downward, but the displacement of the box is horizontal. Therefore, the angle between the gravitational force and displacement is 90 degrees, and the cosine of 90 degrees is 0. So, the work done by the gravitational force is zero.
Work = Force * Distance * Cos(theta) = 0
The work done by the gravitational force is zero Joules.
3. Work done by the normal force:
The normal force acts perpendicular to the surface of contact. Since the box is not moving vertically, the normal force is equal in magnitude but opposite in direction to the gravitational force. Since the displacement is horizontal, the angle between the normal force and displacement is 90 degrees, and the cosine of 90 degrees is 0. Therefore, the work done by the normal force is also zero.
Work = Force * Distance * Cos(theta) = 0
The work done by the normal force is zero Joules.
4. Work done by the frictional force:
The frictional force is in the opposite direction to the applied force and parallel to the displacement. The magnitude of the frictional force can be calculated using the formula:
Frictional Force = Coefficient of friction * Normal force
In this case, the coefficient of kinetic friction is given as 0.19, and the normal force is equal in magnitude but opposite in direction to the gravitational force, which is the weight of the box (mass * acceleration due to gravity). So, the magnitude of the frictional force is:
Frictional Force = 0.19 * (50 kg * 9.8 m/s^2) = 94.05 N
The work done by frictional force can be calculated using the formula:
Work = Force * Distance * Cos(theta)
In this case, the magnitude of the frictional force is 94.05 N, the distance traveled by the box is 7.1 m, and the angle between the frictional force and displacement is 180 degrees. So, the cosine of 180 degrees is -1. Therefore, the work done by the frictional force is:
Work = 94.05 N * 7.1 m * (-1) = -667.155 J
So, the work done by the frictional force is -667.155 Joules.
In summary, the work done on the box by each of the four forces is as follows:
1. Work done by the applied force: 1232.5 J (positive)
2. Work done by the gravitational force: 0 J (zero)
3. Work done by the normal force: 0 J (zero)
4. Work done by the frictional force: -667.155 J (negative)