Question

When a solid sphere is released on a fixed rough inclined plane so that it rolls purely. When it reaches the ground it has the translational kinetic energy of $100J$. What would have been the translational kinetic energy at the bottom, if the plane was smooth? (In $J$) :-

• A. 140.00
• B. 140
• C. 140.0

Answer 1

Answer: (A), (B), (C)

When, $\mu \ne 0$
Work energy theorem –
$PE=(KE{)}_T+(KE{)}_R$
$mgh=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}_c^2$
$mgh=\frac{1}{2}\times \frac{2}{5}m{R}^2{\omega }^2+\frac{1}{2}m{v}_c^2$
$mgh=\frac{1}{2}m{v}_c^2\left[\frac{7}{5}\right]\cdots \left(i\right)$
Given that translation kinetic energy $=100J$
$\therefore m{v}_c^2=200J$
From equation $\left(i\right)$
$mgh=140J$
If the inclined plane was smooth
Then, $\mu =0$
By the energy conservation
Loss in $PE=$ Gain in $KE$
$=\frac{7}{5}\times 100=140\text{J}$
or
For smooth surface, $K_f=U_i=140\text{J}$