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A cheerleader lifts his 37.4 kg partner straight up off the ground a distance of 0.817 m before releasing her. The acceleration of gravity is 9.8 m/s 2. If he does this 27 times, how much work has he done?Answer in units of J

User Zorglube
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1 Answer

2 votes

Answer:

W = 8085.064 J

Step-by-step explanation:

given,

mass of the cheerleaders partner = 37.4 Kg

distance above which she was lift = 0.817 m

acceleration due to gravity = 9.8 m/s²

number of time she was picked = 27 times

work he done = ?

now,

Work done will be equal to the potential energy into number times she was lifted.

Work done = N m g h

W = 27 x 37.4 x 9.8 x 0.817

W = 8085.064 J

work done by his partner is equal to W = 8085.064 J

User Igor Luzhanov
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8.1k points