Question

A particle of mass $2kg$ moves on a smooth horizontal plane under the action of a single force $\vec{F}=(3\hat{i}+4\hat{j})N$. Under this force it is displaced from $(0,0)$ to $(1m,1m)$. The initial speed of the particle is $\sqrt{3}m/s$ and final speed is $\sqrt{5k}m/s$. Find the value of $k$.

Answer 1

Answer: 2

$W=\Delta KE=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2$
Here, $W=\vec{F}\cdot \vec{d}$
$=(3\hat{i}+4\hat{j})\cdot (\hat{i}+\hat{j})$
$=3+4=7J$
$\Rightarrow 7=\frac{1}{2}\times 2\times v_2^2-\frac{1}{2}\times 2\times (\sqrt{3}{)}^2$
$\Rightarrow v_2^2=10$
$\Rightarrow v_2=\sqrt{10}m/s$