A block of weight w = 15.0 N sits on a frictionless inclined plane, which makes

Question

A block of weight w = 15.0 N sits on a frictionless inclined plane, which makes an angle theta = 20.0^circ with respect to the horizontal, as shown in the figure. A force of magnitude F = 5.13 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 W_total 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.

Explanation / Answer

net work done equals the change in KE, since the block moves at a constant velocity, there is no change in KE and no net work done it this means that the work done by the pulling force must be equal in magnitude and opposite in sign to the work done by gravity the work done by gravity = - component of gravitational force down plane x L this component of force is mg sin(theta), so work done by gravity = - mg sin(theta) L thus the work done by F is mg sin(theta)L, and the sum of the two is zero joules So, required answer = 0 J

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