A horizontal force of 5 text{ N} is applied o | vedclass

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(D) Step $1$: Calculate the acceleration using Newton's second law: $a = \frac{F}{m} = \frac{5 	ext{ N}}{5 	ext{ kg}} = 1 	ext{ m/s}^2$.Step $2$: C

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$250 	ext{ J}$

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(D) Step $1$: Calculate the acceleration using Newton's second law: $a = \frac{F}{m} = \frac{5 	ext{ N}}{5 	ext{ kg}} = 1 	ext{ m/s}^2$.Step $2$: Calculate the final velocity after $t = 10 	ext{ s}$ using $v = u + at$: $v = 0 + (1 	ext{ m/s}^2)(10 	ext{ s}) = 10 	ext{ m/s}$.Step $3$: Calculate the change in kinetic energy: $\Delta K = K_f - K_i = \frac{1}{2}mv^2 - 0 = \frac{1}{2} 	imes 5 	ext{ kg} 	imes (10 	ext{ m/s})^2 = \frac{1}{2} 	imes 5 	imes 100 = 250 	ext{ J}$.

$A$ horizontal force of $5 \text{ N}$ is applied on a body of mass $5 \text{ kg}$, which is initially at rest on a frictionless table. What is the change in kinetic energy of the body after $10 \text{ s}$?

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