Question

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

Answer 1

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,

$\Delta P=P_f-P_i$

$\Delta P=W_f-W_i$

$\Delta P=151.44-71.1\times 1.52$

$\Delta P=43.36\ J$