Acheerleader 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

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

this means that the temperature of both substances will be equal

0.19 m/s and 0.45 m/s