Question

A shot put player throws a shot put of mass 3 kg. If it crosses the top of the wall of 2m high at a speed of $4m{s}^{-1}$, compute the total mechanical energy gained by the shot put when it crosses the wall. given, $g=9.8m{s}^{-2}$

Answer 1

Step 1: Data provided,

$$\begin{gathered} Mass\left(m\right)=3kg \\ hight\left(h\right)=2m \\ speedat2mheight\left(v\right)=4m{s}^{-1} \end{gathered}$$

Step 2: Formula used,

$$\begin{gathered} K=\frac{1}{2}m{v}^2,andU=mgh,and \\ T=K+U \end{gathered}$$

Where, K= kinetic energy, U= potential energy, T= total mechanical energy, v = velocity and h= height

Step 3: Calculation,

The potential energy of the shot put when it crosses the wall

$$\begin{gathered} U=mgh \\ =3kg\times 9.8m{s}^{-2}\times 2m \\ =58.8J \end{gathered}$$

The kinetic energy of the shot put when it crosses the wall

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\times 3kg\times {\left(4m{s}^{-1}\right)}^2 \\ =\frac{1}{2}\times 3kg\times \left(4m{s}^{-1}\right)\times \left(4m{s}^{-1}\right) \\ =24J \end{gathered}$$

Total mechanical energy

$$\begin{gathered} T=K+U \\ T=24J+58.8J \\ T=82.8J \end{gathered}$$

Hence, the total mechanical energy gained by the ball is $82.8J$.