5.0 kg block moves in a straight line on a horizontal frictionless surface under

Question

5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. The scale of the figure's vertical axis is set by Fs = 16.0 N. How much work is done by the force as the block moves from the origin tox = 8.0 m?

A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. The scale of the figure's vertical axis is set by Fs = 16.0 N. How much work is done by the force as the block moves from the origin to x = 8.0 m?

Explanation / Answer

workdone=Force*displacement

if force is not constant then we express force in terms of displcement and we integrate F.dX to get workdone

the integration value will equal to area under the curve for the graph F Vs X(displacement)

so in above problem work done by force is area under the curve

area=16*2+16*2/2+0*2-2*8/2

=32+16+0-8

=40

so workdone by force=40Joules

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