A particle of mass $2kg$ moves on a smooth horizontal plane under the action of a single force$\overrightarrow{F}=(3\hat{i}+\hat{j})N$. Under this force it is displaced from $(0,0)$ to $(1m,1m)$. The initial speed of the particle is $\surd 3m/s$ and final speed is $\surd 5km/s$. Find the value of $k$.

Find the work done by given force.
Formula used: Work done$=\overrightarrow{F}\cdot \overrightarrow{s}$
Given,
Mass of the particle $m=2kg$
Force $\overrightarrow{F}=(3iˆ+4jˆ)N$
Displacement $(0,0)$ to $(1m,1m)$
Displacement vector $\hat{i}+\hat{j}$
$W=\overrightarrow{F}\cdot \overrightarrow{s}=(\hat{i}+4\hat{j}).(\hat{i}+\hat{j}=3+4=7J)$
Find the final velocity.
Formula used: $(\text{W}orkdone{)}_{\text{n}etforce}=\Delta K.E$
Apply work energy theorem.
$W==\Delta KE=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2$
$\Rightarrow 7=\frac{1}{2}\times 2\times v_2^2-\frac{1}{2}\times 2\times (\sqrt{3}{)}^2$
$\Rightarrow v_2^2=10\Rightarrow v_2=\sqrt{1}0m/s$
Final answer: $2$