Question

Now consider another situation. A force $\overline{F}=(4\hat{i}+3\hat{j})N$ acts on a particle of mass 2 kg. The particle under the action of this force moves from the origin to a point A (4m, -8m). Initial speed of the particle, i.e., its speed at the origin is $2\sqrt{6}m{s}^{-1}$. Figure 8.288 shows three paths for the motion of the particle from O to A.
Speed of the particle at A will be nearly

• A. 4.0$m{s}^{-1}$
• B. 2.8$m{s}^{-1}$
• C. 3.6$m{s}^{-1}$
• D. 5.6$m{s}^{-1}$

Answer 1

The correct option is A 4.0$m{s}^{-1}$

The work-energy theorem for the situation is expressed as:

$\frac{1}{2}m{v}^2=F.S$

$\frac{1}{2}\times 2\times (4\times 6-v^2)=5\times \sqrt{80}$

$+24-v^2=5\times \sqrt{80}$

$v^2=44-24$

$v^2=20$

$v=\sqrt{20}\Rightarrow v=4.4m/s$