Question

A particle of mass $10\text{kg}$ is moving in a straight line. If its position $x$ with time $t$ is given by $x\left(t\right)=(t^3-2t-10)$ $\text{m}$, then the force acting on it at the end of $4^{th}$ second is

• A. $24\text{N}$
• B. $240\text{N}$
• C. $300\text{N}$
• D. $1200\text{N}$

Answer 1

The correct option is B $240\text{N}$

It is given that the position of the particle is ,
$x=(t^3-2t-10)$
Differentiating the position wrt time we get velocity

$v=\frac{dx}{dt}=3t^2-2$,

and differentiating velocity we get acceleration

$a=\frac{dv}{dt}=6ta=6t$.
$a_{t=4}=24{\text{m/s}}^2$
Now for finding the force acting at $t=4\text{s}$

$\therefore F_{t=4}=m{a}_{t=4}=10\times 24=240\text{N}$

Hence option B is the correct answer