A force of 4 ,N acts on a 10 ,kg body initial | vedclass

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(D) Mass of the body, $m = 10 \,kg$. Force, $F = 4 \,N$. Acceleration produced in the body due to the applied force is $a = \frac{F}{m} = \frac{4}{10}

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$3$

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(D) Mass of the body, $m = 10 \,kg$. Force, $F = 4 \,N$. Acceleration produced in the body due to the applied force is $a = \frac{F}{m} = \frac{4}{10} = 0.4 \,m/s^2$. For the interval $0 \leq t \leq 1 \,s$, the distance traveled $s_1$ is $s_1 = u t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} 	imes 0.4 	imes (1)^2 = 0.2 \,m$. Work done $W_1 = F 	imes s_1 = 4 	imes 0.2 = 0.8 \,J$. For the interval $1 \,s \leq t \leq 2 \,s$, the initial velocity $v_i$ at $t = 1 \,s$ is $v_i = u + a t = 0 + 0.4 	imes 1 = 0.4 \,m/s$. The distance traveled $s_2$ during this interval $(t = 1 \,s)$ is $s_2 = v_i t + \frac{1}{2} a t^2 = 0.4 	imes 1 + \frac{1}{2} 	imes 0.4 	imes (1)^2 = 0.4 + 0.2 = 0.6 \,m$. Work done $W_2 = F 	imes s_2 = 4 	imes 0.6 = 2.4 \,J$. The ratio $\frac{W_2}{W_1} = \frac{2.4}{0.8} = 3$.

$A$ force of $4 \,N$ acts on a $10 \,kg$ body initially at rest. Let $W_1$ be the work done by the force during $0 \leq t \leq 1 \,s$. Likewise, $W_2$ is the work done by the force during $1 \,s \leq t \leq 2 \,s$, where $t$ is time in seconds. The ratio $\frac{W_2}{W_1}$ is

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