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nydimaria [60]
2 years ago
15

A shot-putter puts a shot (weight=71.1N) that leaves his hand at adistance of 1.52m above the ground. (a) Find the work done by

thegravitational force when the shot has risen to a height of 2.13mabove the ground. (b) Determine the change(ΔPE=PEfinal-PEinitial) in the gravitationalpotential energy of the shot.
Need help with B
Physics
1 answer:
dolphi86 [110]2 years ago
4 0

Explanation:

It is given that,

weight of the shot- putter, W = 71.1 kg

Initial position, x_i=1.52\ m

Final position, x_f=2.13\ m

To find,

Work done and the change in potential energy.

Solution,

(a) Let W is the work done by the gravitational force when the shot has risen to a height of 2.13 m above the ground. It is given by :

W_f=F\times x_f

W_f=71.1\times 2.13

W_f=151.44\ J

(b) We know that the potential energy is equal to the work done by an object such that,

\Delta P=P_f-P_i

\Delta P=W_f-W_i

\Delta P=151.44-71.1\times 1.52

\Delta P=43.36\ J

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abruzzese [7]

Answer:

The answer to your question is:   W = 390 J

Explanation:

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mixas84 [53]

Answer:

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