A block of weight w = 20.0 sits on a frictionless inclined plane, which makes an

Question

A block of weight w = 20.0 sits on a frictionless inclined plane, which makes an angle (theta)= 32.0 with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 10.6 , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. A) 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= 4.80m 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. B)What is the work done on the block by the force of gravity as the block moves a distance L= 4.80m up the incline? Express your answer numerically in joules. C)What is the work done on the block by the applied force as the block moves a distance L= 4.80m up the incline? Express your answer numerically in joules. D)What is the work done on the block by the normal force as the block moves a distance L= 4.80m up the inclined plane? Express your answer numerically in joules.

Explanation / Answer

W = force * displacement

a) net force on ball in direction of motion = 0 (as speed is constant)

so, work done by all forces = 0 J

b) force = w*sin(32) = 10.5984 N

so, work done = force*(-4.8)

work = -50.88

c) force = 10.6 N

work = 10.6*(4.8)

work = 50.88 J

d) displacement in direction of force = 0

work done by normal force = 0 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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