A body of mass 5kg is dropped from a tower of height 10m. Find the difference between work done (J) by gravity and change in potential energy of the body from ′A′ to ′B′ (Take g=10ms2)
A
−1000joule
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B
−100joule
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C
1000joule
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D
2000joule
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Solution
The correct option is C1000joule
Work done by gravity from 'A' to 'B' WAB=+mgh;
because gravity is acting downwards and body is also getting displaced downwards.
⇒Now change in potential energy from 'A' to 'B' ΔUA→B=UB−UA...(i)=mghB−mghA=mg(0)−mgh=−mgh
(Where B is the refrence point)
Now, difference between workdone (J) by gravity and change in potential energy of body from ′A′ to ′B′ is given by WA→B−UA→B=(+mgh−(−mgh))=2mgh=2×5×10×10=1000joule