Question

A ball of mass $50\text{g}$ and relative density $0.5$ strikes the surface of water with a velocity of $20\text{m/s}$. It comes to rest at a depth of $2\text{m}$. Find the work done by the resisting force in water.
$($Take, $g=10{\text{m/s}}^2)$

• A. $-6\text{J}$
• B. $-7\text{J}$
• C. $-9\text{J}$
• D. $-8\text{J}$

Answer 1

The correct option is C $-9\text{J}$

From work energy theorem,
$\text{Work done by all the forces}=\Delta KE$

$\Rightarrow W_{\text{gravity}}+W_{\text{resistance}}+W_{\text{buoyancy}}=K{E}_f-K{E}_i$

$\Rightarrow mgh+W_r-2mgh=0-\frac{1}{2}m{v}^2$

$[W_{\text{buoyancy}}=-F_{\text{buoyancy}}h$$=-\rho gVh=-\rho g\times \frac{m}{\sigma }\times h=-\frac{mgh}{\sigma /\rho }$
$=-\frac{mgh}{0.5}=-2mgh]$

$\Rightarrow W_r=-(\frac{1}{2}m{v}^2-mgh)$

$\Rightarrow W_r=-(\frac{1}{2}\times 50\times {10}^{-3}\times {20}^2-50\times {10}^{-3}\times 10\times 2)$

$\Rightarrow W_r=-9\text{J}$