In a shotput event an athlete throws the shotput of mass 10 kg wi… - Vidyadip

MCQ

In a shotput event an athlete throws the shotput of mass $10 \ kg$ with an initial speed of $1 \mathrm{~m} \mathrm{~s}^{-1}$ at $45^{\circ}$ from a height $1.5 m$ above ground. Assuming air resistance to be negligible and acceleration due to gravity to be $10 \mathrm{m} \mathrm{~s}^{-2}$, the kinetic energy of the shotput when it just reaches the ground will be:

A
$20.5J$
B
$5.0J$
C
$52.5J$
✓
$155.0J$

✓

Answer

Correct option: D.

$155.0J$

If air resistance is negligible, total mechanical energy of the system will remain constant.
And let us take ground as a reference where potential energy will be zero.
According to the problem, $\mathrm{h}=1.5 \mathrm{~m}, \mathrm{v}=1 \mathrm{~m} / \mathrm{s}, \mathrm{m}=10 \mathrm{~kg}, \mathrm{~g}=10 \mathrm{~m} \mathrm{~s}^{-2}$
Initial energy of the shotput $=(\mathrm{PE})_{\mathrm{i}}+(\mathrm{KE})_{\mathrm{i}}$
$=\text{mgh}+\frac{1}{2}\text{mv}^2$
$=10\times10\times1.5+\frac{1}{2}\times10\times(1)^2$
$=150+5$
$=155.0\text{J}$
From conservation of mechanical energy,
$\text{(PE)}_\text{i}+\text{(KE)}_\text{i}=\text{(PE)}_\text{f}+\text{(KE)}_\text{f}$
So, final kinetic energy of the shotput is $155J$

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