Starting from rest, a $10 k g$ object is acted upon by only one force as indicated in figure. Find the total work done by the force.

125 J

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245 J

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490 J

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180 J

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Solution

The correct option is D $180 J$
$Change in velocity = \frac{area under F-t graph}{mass}$
$= \frac{40 + (- 10)}{5} = 6 m / s$
Work done by the force = change in kinetic energy
$W = Δ K . E . = \frac{1}{2} (10) 6^{2} = 180 J$

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Starting from rest, a 10 kg object is acted upon by only one force as indicated in figure. Find the total work done by the force.

The correct option is D 180 JChange in velocity=area under F-t graphmass=40+(−10)5=6m/sWork done by the force = change in kinetic energyW=ΔK.E.=12(10)62=180J

Starting from rest, a $10 k g$ object is acted upon by only one force as indicated in figure. Find the total work done by the force.

The correct option is D $180 J$
$Change in velocity = \frac{area under F-t graph}{mass}$
$= \frac{40 + (- 10)}{5} = 6 m / s$
Work done by the force = change in kinetic energy
$W = Δ K . E . = \frac{1}{2} (10) 6^{2} = 180 J$