A block of weight w = 35.0N sits on a frictionless inclined plane, which makes a

Question

A block of weight w = 35.0N sits on a frictionless inclined plane, which makes an angle ? = 30.0? with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 17.5N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.

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 = 2.60m 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.

b)What is Wg, the work done on the block by the force of gravity w?  as the block moves a distance L = 2.60m up the incline?

c)What is WF, the work done on the block by the applied force F?  as the block moves a distance L= 2.60m up the incline?

d)What is WN, the work done on the block by the normal force as the block moves a distance L = 2.60m up the inclined plane?

A block of weight w = 35.0N sits on a frictionless inclined plane, which makes an angle ? = 30.0? with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 17.5N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. 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 = 2.60m 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. b)What is Wg, the work done on the block by the force of gravity w? as the block moves a distance L = 2.60m up the incline? c)What is WF, the work done on the block by the applied force F? as the block moves a distance L= 2.60m up the incline? d)What is WN, the work done on the block by the normal force as the block moves a distance L = 2.60m up the inclined plane?

Explanation / Answer

a) from work -energy theorem ,

total work done = changve in K.E.

its pulled by constant speed.
vchange in K.E> = 0

Net work done= 0

b) by gravity = -mgsin30 x L = -35sin30 x 2.60 = - 45.4 J

c) W = F.L = 17.5 x 2.60 = 45.5 J

d) normal force is perpendicular to dispalcement.

work d>

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