Question

A body of constant mass $m=1kg$ moves under variable force F as shown. If at $t=0,S=0$ and velocity of the body is $\sqrt{20}m/s$ and the force is always along direction of velocity then choose the incorrect options

• A. velocity of the particle will increase upto $S=2m$ and then decrease.
• B. the final velocity at $S=6mis10m/s$
• C. the final velocity at $S=6mis4\sqrt{5}m/s$
• D. the acceleration is constant up to $S=2m$ and then it is negative

Answer 1

The correct option is B the final velocity at $S=6mis10m/s$

$\begin{array}{c}w=\frac{1}{2}\left(6+2\right)10\\ =40J\\ w=\Delta k\\ \Rightarrow 40=\frac{1}{2}m\left[v^2-20\right]\\ \Rightarrow 40=\frac{v^2-20}{2}\\ \Rightarrow 80+20=v^2\\ \Rightarrow v=10m/s\end{array}$

Hence,

option $\left(B\right)$ is correct answer.