Work on a Block Sliding Up a Frictionless Incline A block of weight w = 15.0N si

Question

Work on a Block Sliding Up a Frictionless Incline

A block of weight w = 15.0N sits on a frictionless inclined plane, which makes an angle ? = 30.0? with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 7.50N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.

Part A

The block moves up an incline with constant speed. What is the total work Wtotal done on the block by all forces as the block moves a distance L = 2.70m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest.

Express your answer numerically in joules.

Part B

What is Wg, the work done on the block by the force of gravity w?  as the block moves a distance L = 2.70m up the incline?

Express your answer numerically in joules.

Part C

What is WF, the work done on the block by the applied force F?  as the block moves a distance L= 2.70m up the incline?

Express your answer numerically in joules.

Part D

What is WN, the work done on the block by the normal force as the block moves a distance L = 2.70m up the inclined plane?

Express your answer numerically in joules.

Work on a Block Sliding Up a Frictionless Incline A block of weight w = 15.0N sits on a frictionless inclined plane, which makes an angle ? = 30.0? with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 7.50N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. Part A The block moves up an incline with constant speed. What is the total work Wtotal done on the block by all forces as the block moves a distance L = 2.70m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest. Express your answer numerically in joules. Part B What is Wg, the work done on the block by the force of gravity w? as the block moves a distance L = 2.70m up the incline? Express your answer numerically in joules. Part C What is WF, the work done on the block by the applied force F? as the block moves a distance L= 2.70m up the incline? Express your answer numerically in joules. Part D What is WN, the work done on the block by the normal force as the block moves a distance L = 2.70m up the inclined plane? Express your answer numerically in joules.

Explanation / Answer

Part A:

The block has already started moving at a constant speed. The total work done W on the block is 0 J. Because there is no change in the kinetic energy.

Wtotal=change in kinetic energy

       =0J

Part B:

Wg=-mgy

    =-(15.0N)(2.70m)sin(30)

   =-20.25J

Part C:

Wtotal=Wg+WF

WF=Wtotal-Wg

    =0-(-20.25J)

    =20.25J

Part D:

Since the normal force is perpendicular to the direction of motion, WN is 0J

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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