The figure shows the frictional force versus | vedclass

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(A) The loss in kinetic energy is equal to the work done by the frictional force,which is equal to the area under the force-displacement graph.Area =

Vedclass · Accepted Answer

$250$

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(A) The loss in kinetic energy is equal to the work done by the frictional force,which is equal to the area under the force-displacement graph.Area = Area of the trapezoid from $x = 0$ to $5$ + Area of the rectangle from $x = 5$ to $10$ + Area of the trapezoid from $x = 10$ to $20$.Area = $\frac{1}{2} 	imes (5 + 15) 	imes 5 + (15 	imes 5) + \frac{1}{2} 	imes (15 + 10) 	imes 10$Area = $\frac{1}{2} 	imes 20 	imes 5 + 75 + \frac{1}{2} 	imes 25 	imes 10$Area = $50 + 75 + 125 = 250\, J$.Therefore,the loss in kinetic energy is $250\, J$.

The figure shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s = 0$ to $20\, m$ will be......$J$.

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