A force overrightarrow{F} = 4 hat{i} - 15 hat | vedclass

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(B) According to the work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.$W = \Delta K = K_f -

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) According to the work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.$W = \Delta K = K_f - K_i$Given:Force $\overrightarrow{F} = (4 \hat{i} - 15 \hat{j}) 	ext{ N}$Displacement $\overrightarrow{S} = 6 \hat{i} 	ext{ m}$Initial kinetic energy $K_i = 7 	ext{ J}$Work done $W = \overrightarrow{F} \cdot \overrightarrow{S} = (4 \hat{i} - 15 \hat{j}) \cdot (6 \hat{i}) = (4 	imes 6) + (-15 	imes 0) = 24 	ext{ J}$Using the work-energy theorem:$K_f - K_i = W$$K_f - 7 = 24$$K_f = 24 + 7 = 31 	ext{ J}$Therefore, the kinetic energy at the end of the displacement is $31 	ext{ J}$.

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