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A force of 200 N compresses a steel rod of length 1.5 m and area of cross section 3 mm through a distance of 4 mm. Find the work done. Work done |
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Answer» Solution :WORK done `=(1)/(2)` (Loan) (CHANGE of length) `=(1)/(2) (F)(DeltaL)=(1)/(2) (200) ((4)/(10^(3)))=0.4` Joule, So workdone `=0.4` Joule |
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