A bullet of mass 10 ; g leaves a rifle at an | vedclass

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(D) According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy of the body.Here,the work done against a

Vedclass · Accepted Answer

$3750$

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(D) According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy of the body.Here,the work done against air resistance $(W_{air})$ is equal to the loss in kinetic energy.Mass of the bullet,$m = 10 \; g = 0.01 \; kg$.Initial velocity,$u = 1000 \; m/s$.Final velocity,$v = 500 \; m/s$.Work done against air resistance $W = \Delta K = K_i - K_f = \frac{1}{2} m (u^2 - v^2)$.$W = \frac{1}{2} 	imes 0.01 	imes [(1000)^2 - (500)^2]$.$W = 0.005 	imes [1,000,000 - 250,000]$.$W = 0.005 	imes 750,000$.$W = 3750 \; J$.

$A$ bullet of mass $10 \; g$ leaves a rifle at an initial velocity of $1000 \; m/s$ and strikes the earth at the same level with a velocity of $500 \; m/s$. The work done in joule $(J)$ in overcoming the resistance of air will be:

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