A block of weight w = 25.0 N sits on a frictionless inclined plane, which makes

Question

A block of weight w = 25.0 N sits on a frictionless inclined plane, which makes an angle = 34.0 with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 14.0 N , 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 = 4.70 m 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.

Part B

What is Wg, the work done on the block by the force of gravity w  as the block moves a distance L = 4.70 m up the incline?

Part C

What is WF, the work done on the block by the applied force F  as the block moves a distance L = 4.70 m up the incline?

Part D

What is WN, the work done on the block by the normal force as the block moves a distance L = 4.70 m up the inclined plane?

Explanation / Answer

Here,

W = 25N

theta = 34 degree

F = 14 N

part A)

let the net work done

as the speed is constant

Using work energy theorum

total work done by all forces = change in kinetic energy

total work done by all forces = 0 J

the total work done by all forces is 0 J

part B)

work done by gravity

Wg = -W * L * sin(theta)

Wg = - 25 * 4.70 * sin(34)

Wg = -65.7 J

the work done by the gravity is -65.7 J

part C)

work done by the frictional force = -F * d

work done by the frictional force = -14 * 4.70 J

work done by the frictional force = -65.7 J

part D)

as the normal force is always perpendicular to the direction of motion

Work done by normal force = N * d * sin(90)

Work done by normal force = 0 J

the Work done by normal force is 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.