Question

A body of mass $1kg$ starts moving from rest at $t=0$ , in a circular path of radius $8m$ . Its kinetic energy varies as a function of times as $KE=2t^2$ Joules, where $t$ is in seconds. Then

• A. Tangential acceleration = $4m{s}^{-2}$
• B. Power of all forces at $t=2s$ is $8W$
• C. First-round is completed in $2s$ .
• D. Tangential force at $t=2s$ is $4N$ .

Answer 1

Answer: (B)

$\frac{1}{2}m{v}^2=2t^2$
$v^2=4t^2\Rightarrow v=2t$
$a_{\text{t}}=\frac{\text{d}\text{v}}{\text{d}\text{t}}\Rightarrow a_{\text{t}}=2{\text{m s}}^{-2}$
$F_{\text{t}}=m{a}_{\text{t}}\Rightarrow F_{\text{t}}=2\text{N}$
Power will be inverted only by tangential force
$P=\left({\vec{F}}_t+{\vec{F}}_c\right)\cdot \vec{V}$ since $F_c\perp V\&F_t\Vert V$
$P=F_{t}V\Rightarrow P=2\times 2t$
$\Rightarrow P=4t\Rightarrow P=8W$