Question

A rope exerts an 18-N force while lowering a 20-kg crate down a plane inclined at 20∘ (the rope is parallel to the plane). A 24-N friction force opposes the motion. The crate starts at rest and moves 11 m down the plane.

Part A:  Determine the final speed of the crate. The system consists of the crate, the surface of the incline and Earth.

Part B: Determine the change in gravitational energy.

Part C: Determine the change in kinetic energy.

Part D: Determine the change in internal energy.

Part E: Determine the work done by the rope.

Answer 1

Given, mass of the crate (m) = 20 kg

Tension in the rope (T) = 18 N.

Given kinetic frictional force (f) as 24 N

From the free body diagram, we can write the following equation for motion along the incline.

m * g * sin(20) - f - T = m * a

=> a = 9.81 * sin(20) - (24 / 20) - (18 / 20)

=> a = 1.2552 m/s² down the incline.

Part A:

From, the equation of linear motion, we have

v² - u² = 2 * a * s

Given, s = 11 m and u = 0 m/sec ; We found, a = 1.2552 m/s² ​

=> v = sqrt(2 * 1.2552 * 11) = 5.26 m/sec

Final speed of the crate = 5.26 m/sec

Part B:

Change in gravitation potential energy = m * g * (h)

Where h is the vertical distance slided.

Therefore, h = 11 * sin(20) = 3.7622 meters

m * g * h = 20 * 9.81 * 3.7622 = 738.15 Joules

Hence, Change in Gravitational Energy = 738.15 Joules

Part C:

Change in kinetic energy = Final kinetic energy - Initial kinetic energy = 0.5 * m * v² - 0.5 * m * u²

Change in KE = 0.5 * 20 * 5.26 * 5.26 - 0 = 276.15 Joules

Therefore, Change in kinetic energy = 276.15 Joules

Part D:

As, the temperature changes of the crate is not given, I am assuming the internal energy remained constant.

Therefore, Change in internal energy is 0

Part E:

Work done = Force * displacement = 18 * 11 = 198 J

Therefore, work done by the rope = 198 J