Question

A small body of mass $m=0.5kg$ is allowed to slide on an inclined frictionless track from rest position as shown in the figure. $(g=10m/s^2)$
If $h$ is double of that minimum height required to complete the loop sucessfully, calculate resultant force on the block at position $H$ in newton

Answer 1

Using principle of conservation of energy, we have

$mg{h}_{min}=\frac{1}{2}m{v}^2$

where v is the velocity of the body at the lowest point of the circular part of the track.

We know that to complete the loop successfully, $v\ge \sqrt{5rg}$

$\Rightarrow mg{h}_{min}=\frac{1}{2}m(\sqrt{5rg}{)}^2$
$\Rightarrow mg{h}_{min}=\frac{5}{2}mrg$
$\Rightarrow h_{min}=\frac{5}{2}r$

Now, given

$h=2\times h_{min}=5r$

Applying principle of conservation of energy at position H, we have

$mgh-mg\left(2r\right)=\frac{1}{2}m{v}^2$

$\Rightarrow mg\left(5r\right)-mg\left(2r\right)=\frac{1}{2}m{v}^2$

$\Rightarrow 3mgr=\frac{1}{2}m{v}^2$

$\Rightarrow v^2=6rg$

Now at position H,

$F_{resultant}=\frac{m{v}^2}{r}-mg$

$F_{resultant}=\frac{6mrg}{r}-mg$

$F_{resultant}=6mg-mg=5mg$

Hence, the answer is $5mg$.