Question

The work done to accelerate a body from $30m{s}^{-1}$ to $60m{s}^{-1}$ is three times the work done to accelerate it from $10m{s}^{-1}$ to 'v'. The value of 'v' in $m{s}^{-1}$:

• A. 30
• B. $20\sqrt{2}$
• C. $30\sqrt{3}$
• D. $10\sqrt{10}$

Answer 1

The correct option is B $20\sqrt{2}$

: $30m{s}^{-1}to60m{s}^{-1}$

$\&$

$10m{s}^{-1}tovm{s}^{-1}$

$\Delta k=\frac{m{v_r}^2}{2}-\frac{m{v_1}^2}{2}$

$\Delta k_{30to60}=\frac{m{\left(60\right)}^2}{2}-\frac{m{\left(30\right)}^2}{2}$

$\Delta k_{10tov}=\frac{m{v}^2}{2}-\frac{m{\left(10\right)}^2}{2}$

$\frac{\Delta k_{30to60}}{\Delta k_{10tov}}=3$

$\frac{m\left(\frac{{60}^2}{2}-\frac{{30}^2}{2}\right)}{m\left(\frac{v^2}{2}-\frac{{10}^2}{2}\right)}=3$

$v=20\sqrt{2}m{s}^{-1}$