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A particle of mass m = 1 kg is moving in space in X direction with a velocity of 10 ms–¹. A 4 N force acting in Y direction is applied on it for a time interval of 5.0 s. Later a 5 N force was applied on it in Z direction for 4.0 s
Calculate the total work done by both the force

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a

200 J

b

400 J

c

300 J

d

600 J

answer is B.

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Detailed Solution

$Δ \vec{p} = Impulse m \vec{V} - m \vec{u} = 4 \times 5 \hat{j} + 5 \times 4 \hat{k} 1 \times \vec{V} - 1 \times 10 \hat{i} = 20 \hat{j} + 20 \hat{k} \vec{V} = 10 \hat{i} + 20 \hat{j} + 20 \hat{k} ∴ V = \sqrt{10^{2} + 20^{2} + 20^{2}} = 30 {ms}^{- 1}$

Work energy theorem gives–
$W = \frac{1}{2} {mV}^{2} - \frac{1}{2} {mu}^{2} = \frac{1}{2} \times 1 \times (30^{2} - 10^{2}) = 400 J$

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