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Physics Question on types of forces

The force acting on a body of mass 10 kg is (2i^+j^ −k^)(2\mathbf{\hat{i}}+\mathbf{\hat{j}}\,-\mathbf{\hat{k}}) N. If the body is initially at rest. The velocity at the end of the 20 sec is:

A

232\sqrt{3}

B

323\sqrt{2}

C

626\sqrt{2}

D

262\sqrt{6}

Answer

262\sqrt{6}

Explanation

Solution

Here: mass of the body m = 10 kg Force (F→)=(2i^+j^−k^)N(\overrightarrow{F})=(2\hat{i}+\hat{j}-\hat{k})N Initially velocity of the body u=0u=0 time t = 20 sec Acceleration a→=F→m=(2i^+j^−k^)10\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{(2\hat{i}+\hat{j}-\hat{k})}{10} Hence final velocity of the body v→=u+a→ t\overrightarrow{v}=u+\overrightarrow{a}\,t =0+(2i^+j^−k^10)×20=0+\left( \frac{2\hat{i}+\hat{j}-\hat{k}}{10} \right)\times 20 =4i^ +2j^−2k^=4\hat{i}\,+2\hat{j}-2\hat{k} Magnitude of final velocity =16+4+4=\sqrt{16+4+4} =24=26=\sqrt{24}=2\sqrt{6}