chp. 7 #19 A block of weight omega = 15.0 N sits on a frictionless inclined plan

Question

chp. 7 #19

A block of weight omega = 15.0 N sits on a frictionless inclined plane, which makes an angle theta = 33.0degee with respect to the horizontal, as show in the figure. A force of magnitude F = 8.17 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 = 3.20 mk 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. What is W_g the work done on the block by the force of gravity omegavector as the block moves a distance L = 3.20 m up the incline? What is W_f the work done on the block by the applied force vector F as the block moves a distance L = 3.20 m up the incline?

Explanation / Answer

PART A) Work done by all forces = change in kinetic energy

= 0 J as speed is constant

PART c) WF = F.L

= 8.17*3.20

= 26.144 J......PART C, not B

PART B) since work done by all forces = 0

Wg + WF = 0

Wg = - WF

= - 26.144 J

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