A shot-putter puts a shot (weight = 70.9 N) that leaves his hand at distance of1

Question

A shot-putter puts a shot (weight = 70.9 N) that leaves his hand at distance of1.60 m above the ground. (a) Find the work done by the gravitation forcewhen the shot has risen to a height of 2.09 m above the ground. Include the correct signfor work.
1 J

(b) Determine the change (PE = PEf -PE0) in the gravitational potential energy of theshot.
2 J (a) Find the work done by the gravitation forcewhen the shot has risen to a height of 2.09 m above the ground. Include the correct signfor work.
1 J

(b) Determine the change (PE = PEf -PE0) in the gravitational potential energy of theshot.
2 J

Explanation / Answer

so according to the work energy theorem, the answer to part a and bare the same. work = F*d          =(70.9)(2.09-1.6)          = 34.741J PE final = mgh            = (70.9)(2.09)            = 148.181 J PE Initial = mgh              = (70.9)(1.6)              = 113.44 J PE = PE final -PE initial = 148.181 - 113.44 = 34.741J Hopefully this helps!

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