Questions
A person of mass 70 kg jumps from a stationary helicopter with the parachute open. As he falls through 50 m height, he gains a speed of . The work done by the viscous air drag is
detailed solution
Correct option is B
From work-energy theorem, net work done by all forces (internal and external) = change in kinetic energy. ⇒ Work done by gravity + work done by air drag=12×70(202−02) ⇒Work done by air drag =14000 - 35000 = -21000 JTalk to our academic expert!
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The work done in moving a body of mass 4 kg with uniform velocity of 5 for 10 seconds on a surface of = 0.4 is (take g = 9.8 )
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