Question

the diagram shows a crate with mass m = 33.9 kg being pushed up an incline that makes an angle Phi = 23.2 degrees with the horizontal. the pushing force is horizontal with magnitude P, and the coefficient of kinetic friction between the crate and the incline is Mu = 0.25. consider the work done on the crate as it moves a distance d = 5.23 m at constant speed. what is work in joules done by the pushing force

Answer 1

The work done by the pushing force is calculated step by step by determining the forces involved, finding the pushing force, and applying the work formula. The final result is approximately 887.68 Joules.

Certainly, let's go through the step-by-step calculations for the work done by the pushing force on the crate as it moves up the incline.

Given data:

Mass of the crate (m) = 33.9 kg

Incline angle (phi) = 23.2 degrees

Coefficient of kinetic friction (mu) = 0.25

Distance moved (d) = 5.23 m

Determine the Components of Forces:

Gravitational force (mg) = 33.9 kg * 9.8 m/s^2 (downward)

Normal force (N) = 33.9 kg * 9.8 m/s^2 * cos(23.2 degrees) (perpendicular to the incline)

Frictional force (f) = mu * N = 0.25 * 33.9 kg * 9.8 m/s^2 * cos(23.2 degrees) (opposite to the direction of motion)

Determine the Net Force in the Horizontal Direction:

P - f = 0 (since the crate moves at constant speed)

P = f

Calculate the Pushing Force (P):

P = 0.25 * 33.9 kg * 9.8 m/s^2 * cos(23.2 degrees)

Use the Work Formula (W = Pd*cos(phi)):

W = 0.25 * 33.9 kg * 9.8 m/s^2 * cos(23.2 degrees) * 5.23 m

Perform the Calculation:

W ≈ 887.68 Joules

So , the work done in joules by pushing force would be 887,68 joules.