Question

Position of a $10\text{kg}$ block moving under influence of a force is given as $s=6t^3+5t^2+10\left(\text{m}\right)$, where $\text{t}$ is in seconds. Find out the force acting after $10sec$.

• A. $5000\text{N}$
• B. $4000\text{N}$
• C. $3600\text{N}$
• D. $3700\text{N}$

Answer 1

The correct option is D $3700\text{N}$

We have, $s=6t^3+5t^2+10$ and we know
Velocity, $v=\frac{ds}{dt}=\frac{d}{dt}(6t^3+5t^2+10)$
$v=18t^2+10t\left(\text{m/s}\right)$
Acceleration, $a=\frac{dv}{dt}=\frac{d}{dt}(18t^2+10t)$
$a=36t+10\left({\text{m/s}}^2\right)$
$a(t=10\text{s})=36t+10=36\times 10+10$
$a(t=10\text{s})=370{\text{m/s}}^2$
Force at $(t=10s)$ = $ma$
$=10\times 370$
$=3700\text{N}$