Question

A body of mass $2\text{kg}$ slide down a curved track which is quadrant of a circle of radius $2\text{m}$. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is
$(g=10{\text{m/s}}^2)$

• A. $5.18\text{m/s}$
• B. $6.32\text{m/s}$
• C. $7\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is B $6.32\text{m/s}$

According to law of conservation of energy
Total mechanical energy = constant.
Initial potential energy $\left(U_i\right)+$Initial kinetic energy $\left(K_i\right)=$ Final potential energy $\left(U_f\right)+$Final kinetic energy $\left(K_f\right)$

$\because$ Initial velocity $=0,K_i=0$
$U_i=mg\times 2$
$U_f=0$ [$\because$ base is take as reference]
$K_f=?$
$\Rightarrow mg\times 2=\frac{1}{2}m\times v_f^2$
$v_f=\sqrt{4g}=\sqrt{40}\text{m/s}=6.32\text{m/s}$
Speed of a body at the bottom of the track is $v_f=6.32\text{m/s}$