Item 2 Part A A block of weight w = 40.0 N sits on a frictionless inclined plane

Question

Item 2 Part A A block of weight w = 40.0 N sits on a frictionless inclined plane, which makes an angle theta = 28.0 with respect to the horizontal, as shown in the figure. (Figure needed to start the block moving from rest 1) A force of magnitude F = 18.8 N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. The block moves up an incline with constant speed. What is the total work done on the block by all forces as the block moves a distance L = 3.20 m up the incline? Include only the work done after the block has started moving at constant speed, not the work 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 = 3.20 m up the incline? Express your answer numerically in joules.

Explanation / Answer

Part A) Since there is no friction, and the block is moving at constant speed (no change in kinetic energy), the total work done on the block is Wnet=0 J.

Part B) the work done by the force of gravity is given by:

Wg= -Fg*Lsin() (negative because gravity is against the direction of motion)

Wg=-40.0N*3.20m*sin(28.0º)= -60.09 J

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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