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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

Answer :

Answer:

W = 8085.064 J

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

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