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Question

Physics Question on laws of motion

A 0.2 kg object at rest is subjected to a force (0.3i^−0.4j^)\left(0.3\hat{i}-0.4\hat{ j}\right) N. What is the velocity after 6 s?

A

(9i^−12j^)\left(9\hat{i}-12\hat{ j}\right)

B

(8i^−16j^)\left(8\hat{i}-16\hat{ j}\right)

C

(12i^−9j^)\left(12\hat{i}-9\hat{ j}\right)

D

(16i^−8j^)\left(16\hat{i}-8\hat{ j}\right)

Answer

(9i^−12j^)\left(9\hat{i}-12\hat{ j}\right)

Explanation

Solution

Here, m=0.2 kg,u=0m = 0.2 \,kg, u = 0 F⃗=(0.3i^−0.4j^)\vec{F}=\left(0.3 \hat{i}-0.4 \hat{j}\right) v⃗\vec{v} = ?, t=6 st = 6 \,s a⃗=F⃗m=(0.3i^−0.4j^)0.2=(32i^−2j^)\vec{a}=\frac{\vec{F}}{m}=\frac{\left(0.3 \hat{i}-0.4 \hat{j}\right)}{0.2}=\left(\frac{3}{2} \hat{i}-2 \hat{j }\right) From v⃗=u⃗+a⃗t\vec{v}=\vec{u}+\vec{a} t v⃗=0+(32i^−2j^÷)×6=9i^−12j^\vec{v}=0+\left(\frac{3}{2} \hat{i}-2 \hat{j }\div \right)\times6=9 \hat{i}-12 \hat{j}