Question

If the ground is sufficiently rough to ensure rolling, what is the kinetic energy of the body now in the given time interval of $2s$?

• A. $18.75J$
• B. $16.67J$
• C. $5.55J$
• D. Cannot be found

Answer 1

The correct option is A $18.75J$

$a=\frac{f}{m}$

$\alpha =\frac{FR-f\left(2R\right)}{I}$

For pure rolling - $a=\alpha R$

$\frac{f}{m}=\left(\frac{FR-f\left(2R\right)}{I}\right)R$

$\Rightarrow \frac{f}{2}=\left(\frac{10\times 1-2f}{4}\right)$

$\Rightarrow f=5-f$

$\Rightarrow 2f=5\Rightarrow f=2.5N$

$a=\frac{f}{m}=\frac{2.5}{2}=1.25m/s^2$

$\alpha =\frac{(10\times 1)-\left(2.5\right)\left(2\right)}{4}=\frac{5}{4}{s}^{-2}$

$V_f=V_i+at$ and $w_f=w_i+\alpha t$

$V_f=0+\left(1.25\right)\times \left(2\right)$ $w_f=0+\frac{5}{4}\times 2$

$V_f=2.5m/s$ $w_f=2.5s^{-1}$

$KE=\frac{1}{2}{mv}^2+\frac{1}{2}{Iw}^2$

$=\frac{1}{2}\times 2\times {\left(2.5\right)}^2+\frac{1}{2}\times 4\times {\left(2.5\right)}^2$

$=18.75$

I am getting as option (A).

Option (B).