A rope exerts an 18N force while lowering a 20 kg crate own a plane inclined at

Question

A rope exerts an 18N force while lowering a 20 kg crate own a plane inclined at 25 degree (the rope is parallel to the plane). A 24N friction force opposes the motion. The crate starts at rest and moves 15 m down the plane. 1) Determine the final speed of the crate. The system son sista of tehe crate, the surface of the incline and Earth. 2) Determine the Change in gravitational energy. 3) Determine the change in kinetic energy. 4) Determine the change in internal energy. 5) Determine the work done by the rope. A rope exerts an 18N force while lowering a 20 kg crate own a plane inclined at 25 degree (the rope is parallel to the plane). A 24N friction force opposes the motion. The crate starts at rest and moves 15 m down the plane. 1) Determine the final speed of the crate. The system son sista of tehe crate, the surface of the incline and Earth. 2) Determine the Change in gravitational energy. 3) Determine the change in kinetic energy. 4) Determine the change in internal energy. 5) Determine the work done by the rope. 1) Determine the final speed of the crate. The system son sista of tehe crate, the surface of the incline and Earth. 2) Determine the Change in gravitational energy. 3) Determine the change in kinetic energy. 4) Determine the change in internal energy. 5) Determine the work done by the rope.

Explanation / Answer

Since the crate is moving down the inclined plane, there are three forces that affect its acceleration. The component of its weight that is parallel to the inclined plane causes it to accelerate. The 18 N force and the friction force cause it to decelerate.

Force parallel = 20 * 9.8 * sin 25 = 196 * sin 25
This is approximately 82.8296 N.
Net force = 196 * sin 25 – 18 – 24 = 196 * sin 25 – 42

a = (196 * sin 20 – 42) ÷ 20 = 9.8 * sin 25 – 2.1
This is approximately 2.04148 m/s^2. To determine the velocity at the bottom of the incline, use the following equation.

vf^2 = vi^2 + 2 * a * d, vi = 0
vf^2 + 2 * (9.8 * sin 25 – 2.1) * 15 = 294 * sin 25 – 63
vf = (294 * sin 25 – 63)
This is approximately 7.8258 m/s.

Part B - Determine the change in gravitational energy.

PE = 20 * 9.8 * h = 196 * h
h = 15 * sin 25
PE = 2940 * sin 25
This is approximately 1242.444 J.

Part C - Determine the change in kinetic energy.

KE = ½ * 20 * (294 * sin 25 – 63) = 2940 * sin 25 – 630
This is approximately 612.444 J

Part D - Determine the change in internal energy.
The change in internal energy is equal to the work that is done by the friction force.
Friction work = 24 * 15 = 360 N * m

Part E - Determine the work done by the rope.

Work = T * d = 18 * 15 = 270 N * m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.