A particle of mass 50 , g moves on a straight | vedclass

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(B) Given mass $m = 50 \, g = 0.05 \, kg$.From the velocity-time graph,the acceleration $a$ is the slope of the line.At $t = 2 \, s$,the slope is $a =

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$0.25 \, N$ along motion,zero,$0.25 \, N$ opposite to motion

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(B) Given mass $m = 50 \, g = 0.05 \, kg$.From the velocity-time graph,the acceleration $a$ is the slope of the line.At $t = 2 \, s$,the slope is $a = \frac{15 - 0}{3 - 0} = 5 \, m/s^2$. The force is $F = ma = 0.05 	imes 5 = 0.25 \, N$ in the direction of motion.At $t = 4 \, s$,the velocity is constant,so acceleration $a = 0$. Thus,$F = 0$.At $t = 6 \, s$,the slope is $a = \frac{0 - 15}{8 - 5} = -5 \, m/s^2$. The force is $F = ma = 0.05 	imes (-5) = -0.25 \, N$. The negative sign indicates the force is opposite to the direction of motion.

$A$ particle of mass $50 \, g$ moves on a straight line. The variation of speed with time is shown in the figure. Find the force acting on the particle at $t = 2, 4,$ and $6 \, s$.

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