A 10 kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20 degrees with the horizontal. If the coefficient of kinetic friction is .40 and the crate is pulled a distance of 5 m, (a) how much work is done by gravitational force? (b) How much work is done by the 100 N force? (c) What is the change in kinetic energy of the crate? (d) What is the speed of the crate after it is pulled 5m?

To answer these questions, we need to use the concepts of work, force, and energy.

(a) To calculate the work done by the gravitational force, we need to determine the component of the gravitational force parallel to the incline. The gravitational force can be calculated using the formula F_gravity = m * g, where m is the mass of the crate and g is the acceleration due to gravity.

Given that the mass of the crate is 10 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the force of gravity: F_gravity = 10 kg * 9.8 m/s^2 = 98 N.

The work done by gravity can be calculated using the formula W_gravity = F_gravity * d * cos(theta), where d is the distance and theta is the angle between the direction of the force and the displacement.

Since the crate is being pulled up the incline, the angle between the gravitational force and the displacement is 180 degrees. Hence, cos(180) = -1.

Therefore, W_gravity = 98 N * 5 m * cos(180) = -490 J. The negative sign indicates that the work done by gravity is in the opposite direction of the displacement.

(b) To calculate the work done by the 100 N force, we can use the formula W_force = F_force * d * cos(theta), where F_force is the applied force, d is the distance, and theta is the angle between the direction of the force and the displacement.

The force applied parallel to the incline is 100 N, and the distance is 5 m. The angle between the applied force and the displacement is 20 degrees, given in the question.

Therefore, W_force = 100 N * 5 m * cos(20) ≈ 866.03 J.

(c) To find the change in kinetic energy of the crate, we can use the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy.

The net work done is the sum of the work done by gravity and the work done by the applied force:

Net work = W_gravity + W_force = -490 J + 866.03 J ≈ 376.03 J.

Since work is equal to the change in kinetic energy, the change in kinetic energy of the crate is equal to the net work done.

Therefore, the change in kinetic energy is approximately 376.03 J.

(d) To calculate the final speed of the crate after it is pulled 5 m, we can use the equation for the work-energy theorem:

Net work = ΔKE = (1/2) * m * (vf^2 - vi^2),

where ΔKE is the change in kinetic energy, m is the mass of the crate, vf is the final velocity, and vi is the initial velocity.

Rearranging the equation, we get:

vf^2 = vi^2 + (2 * ΔKE) / m.

Substituting the given values, we have:

vf^2 = (1.5 m/s)^2 + (2 * 376.03 J) / 10 kg.

Solving this equation, we find:

vf^2 = 2.25 + 75.206 ≈ 77.456,

vf ≈ √77.456 ≈ 8.79 m/s.

Therefore, the speed of the crate after it is pulled 5 m is approximately 8.79 m/s.

87.00m/s