Question

A force $F$ is applied to a cart of mass $10kg$ initially at rest. The quadratic variation of force with time is shown in the graph. The speed of cart at $t=5\text{sec}$ is

• A. $10\text{m/s}$
• B. $8.33\text{m/s}$
• C. $2\text{m/s}$
• D. zero

Answer 1

The correct option is B $8.33\text{m/s}$

Mass of cart $=10\text{kg}$
The equation of force acting on the cart $=2t^2$ [from graph]
Equation of $F-t$ curve (parabolic), $F\left(t\right)=k{t}^2$
Since at $t=1\text{sec},F=2\text{N}\Rightarrow k=2$
$\Rightarrow F\left(t\right)=2t^2$
Acceleration, $a\left(t\right)=\frac{F\left(t\right)}{m}$
$a=\frac{2t^2}{10}{\text{m/s}}^2$
$a=\frac{t^2}{5}{\text{m/s}}^2$
Since acceleration, $a=\frac{dv}{dt}$
$v\left(t\right)=\int a\left(t\right)dt$
$v\left(t\right)=\int \frac{t^2}{5}dt$
$v\left(t\right)=\frac{t^3}{15}+C$
As the cart is initially stationary
$\therefore v\left(0\right)=0$
$\Rightarrow C=0$
$\therefore v\left(t\right)=\frac{t^3}{15}\text{m/s}$
$\Rightarrow$ The speed of cart at $t=5\text{sec}$ is
$v=\frac{5^3}{15}\text{m/s}$
$=\frac{125}{15}\text{m/s}$
$=\frac{25}{3}\text{m/s}$
$v=8.33\text{m/s}$