A 3.90 kg block is pushed 2.80 m up a vertical wall with constant speed by a con

Question

A 3.90 kg block is pushed 2.80 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of Theta=30.0 degree with the horizontal, as shown in the figure below If the coefficient of kinetic friction between block and wall is 0.300, determine the following (a) the work done by F. (b) the work done by the force of gravity (c) the work done by the normal force between block an wall (d) By how much does the gravitational potential energy increase during the block?s motion?

Explanation / Answer

first list all the forces acting on the block.
1) F at angle theta
2) m*g vertically downward
3) friction force, 0.3*normal force, vertically downward

now normal force=F*cos theta
and as velocity is constant., acceleration is 0
net force vertically =0
so F*sin(30)=m*g+0.3*F*cos(30)

F=159.1225

a)work done=force *distance*cos(30)=385.8514 J

b)work done by gravity=3.9*9.8*cos(180)*2.8=-107.016 J

c) angle between normal force and direction of movement=90 degrees

hence work done=0

d)gravitational potential energy increase by 3.9*9.8*2.8= 107.016 J

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