Question

A 0.2 kg object at rest is subjected to a force $\left(\text{0.3} \hat{ i } - \text{0.4} \hat{ j ⁡}\right)$ newton. The velocity vector after 6s is

• A. $\left(9 \hat{ i } - 1 2 \hat{ j ⁡}\right)$
• B. $\left(8 \hat{ i } - 1 6 \hat{ j ⁡}\right)$
• C. $\left(1 2 \hat{ i } - 9 \hat{ j ⁡}\right)$
• D. $\left(1 6 \hat{ i } - 8 \hat{ j ⁡}\right)$

Answer 1

Answer: (A)

Here, m = 0.2 kg ; u = 0
$\overset{→}{\text{F}} = \left(\text{0.3} \hat{\text{i}} - \text{0.4} \hat{\text{j}}\right) \text{N}$
$\overset{→}{\text{v}} = \text{?} \text{, } \text{t} = 6 \text{s}$
$\overset{→}{\text{a}} = \frac{\overset{→}{\text{F}}}{\text{m}} = \frac{\left(\text{0.3} \hat{\text{i}} - \text{0.4} \hat{\text{j}}\right)}{\text{0.2}} = \frac{3}{2} \hat{\text{i}} - 2 \hat{\text{j}}$
From $\overset{→}{\text{v}} = \overset{→}{\text{u}} + \overset{→}{\text{a}} \text{t} \text{, } \overset{→}{\text{v}} = 0 + \left(\frac{3}{2} \hat{\text{i}} - 2 \hat{\text{j}}\right) \times 6$
$= 9 \hat{\text{i}} - 1 2 \hat{\text{j}}$