The length of a smooth inclined plane of inclination 30 degree is 5m. The work done in moving a 10 kg mass from the bottom of the inclined plane to top is
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5
F - mgsin30 = mg
F = 150N
W= 150 × 5
= 750J
F = 150N
W= 150 × 5
= 750J
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0
Answer:
250J
Explanation:
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