A block of weight w = 30.0 N sits on a frictionless inclined plane, which makes
ID: 1396355 • Letter: A
Question
A block of weight w = 30.0 N sits on a frictionless inclined plane, which makes an angle ? = 21.0 ? with respect to the horizontal, as shown in the figure.(Figure 1) A force of magnitude F = 10.8 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 = 2.50 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.
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 = 2.50 m 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 = 2.50 m 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 = 2.50m up the inclined plane?
Express your answer numerically in joules.
Explanation / Answer
part A )
the block is moving with constant velocity ,
the change in kinetic energy of block is zero.
using work energy theorum
change in kinetic energy = net work done
net work done = 0
the net work done on the block is 0
B)
work done by gravity = - W*L*sin(21)
work done by gravity = -30 * 2.50 *sin(21)
work done by gravity = - 27 J
the work done by gravity is -27 J
C)
work done by force F = F * L
work done by force F = 10.8 * 2.5
work done by force F = 27 J
the work done by force F is 27 J
D)
as the normal force is always perpendicular to direction of motion
work done by normal = N * L * cos(90)
work done by normal = 0
work done by normal is 0 J
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