A particle of mass 0.01 ; kg travels with vel | vedclass

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(D) According to the work-energy theorem,the work done on a particle is equal to the change in its kinetic energy.$W = \Delta K = \frac{1}{2} m (v_2^2

Vedclass · Accepted Answer

$0.96$

Vedclass · Answer

(D) According to the work-energy theorem,the work done on a particle is equal to the change in its kinetic energy.$W = \Delta K = \frac{1}{2} m (v_2^2 - v_1^2)$Initial velocity vector: $\vec{v}_1 = 4 \hat{i} + 16 \hat{k} \; m s^{-1}$.Initial speed squared: $v_1^2 = |\vec{v}_1|^2 = 4^2 + 16^2 = 16 + 256 = 272 \; m^2 s^{-2}$.Final velocity vector: $\vec{v}_2 = 8 \hat{i} + 20 \hat{j} \; m s^{-1}$.Final speed squared: $v_2^2 = |\vec{v}_2|^2 = 8^2 + 20^2 = 64 + 400 = 464 \; m^2 s^{-2}$.Given mass $m = 0.01 \; kg$.Substituting the values into the work-energy equation:$W = \frac{1}{2} 	imes 0.01 	imes (464 - 272)$$W = 0.005 	imes 192$$W = 0.96 \; J$.

$A$ particle of mass $0.01 \; kg$ travels with velocity given by $4 \hat{i} + 16 \hat{k} \; m s^{-1}$. After some time,its velocity becomes $8 \hat{i} + 20 \hat{j} \; m s^{-1}$. The work done on the particle during this interval of time is ...... $J$.

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