A body of mass $10 \mathrm{~kg}$ is acted upon by a force given by equation $F=\left(3 t^2-30\right)$…
A body of mass $10 \mathrm{~kg}$ is acted upon by a force given by equation $F=\left(3 t^2-30\right)$ newtons. The initial velocity of the body is $10 \mathrm{~m} / \mathrm{s}$. The velocity of the body after $5 \mathrm{~s}$ is
$4.5 \mathrm{~m} / \mathrm{s}$
$6 \mathrm{~m} / \mathrm{s}$
$7.5 \mathrm{~m} / \mathrm{s}$
$5 \mathrm{~m} / \mathrm{s}$
Solution
Body is acted upon by a force given by equation
$
\begin{aligned}
& F=\left(3 t^2-30\right) \mathrm{N} \\
& \therefore m(v-u)=\int F d t \\
& =\frac{3 t^3}{3}-30 t=t^3-30 t \\
& =5^3-30(5)=125-150=-25 \\
& \therefore \quad m[v-10]=-25 \\
& \Rightarrow \text { Final velocity } \quad v=7.5 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$
.
Asked in: JEE Mains - Motion In One Dimension - Test 3