Question

A force F is applied to the initially stationary cart. The variation of force with time is shown in the figure. The speed of cart at t = 5 sec is

• A. 10 m/s
• B. 8.33 m/s
• C. 2 m/s
• D. Zero

Answer 1

The correct option is B

8.33 m/s

$t^2=\text{K}\times \text{F}$; At t = 5 sec, F = 50 N

$\therefore 5^2=\text{K}\left(50\right)\Rightarrow \text{K}=\frac{1}{2}$

Hence ${\text{t}}^2=\frac{1}{2}\text{F}$ or ${\text{t}}^2=\frac{1}{2}\text{ma}$

or $\text{a}=\frac{2t^2}{m}$ or $\text{a}=\frac{t^2}{5}[\because \text{m}=10\text{kg}]$

or $\frac{dv}{dt}=\frac{t^2}{5}$ or ${\int }_0^{v}dv={\int }_0^5\frac{t^2}{5}dt$

$\therefore v=8.33\text{m}/\text{s}$