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Starting at rest , `a 10kg` object is acted upon by only one force as indicated in figure. Then the total work done by the force is `:`
image
A. `90J`
B. `125J`
C. `245J`
D. `490J`

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Correct Answer - B
Change in velocity `= ( area underF-T graph)/(mass )`
`=(60+(-10))/(10)=5m//s`
`W_(F)=DeltaK.E. =(1)/(2)(10)5^(2)=125J`

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