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4-1.Newton's Laws of Motion
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A particle of mass $50$ gram moves on a straight line. The variation of speed with time is shown in figure. find the force acting on the particle at $t =2,4$ and $6$ seconds.
A$0.25\,N$ along motion, zero,$0.25$ along to motion
B$0.25\,N$ along motion, zero,$0.25$ opposite to motion
C$0.25\,N$ opposite motion, zero,$0.25$ along to motion
D$0.25\,N$ opposite motion, zero, $0.25$ opposite to
Solution
At $t=2\,sec$
$a=\frac{10}{2}=5\,m / s ^2$
So, $F=m a=\frac{50}{1000} \times 5=0.25\,N$
At $t=4\,sec$
$a=0$
So $F=0$
At $t=6\,sec$,
$a =-5\,m / s ^2$
$F =-0.25\,N$
$a=\frac{10}{2}=5\,m / s ^2$
So, $F=m a=\frac{50}{1000} \times 5=0.25\,N$
At $t=4\,sec$
$a=0$
So $F=0$
At $t=6\,sec$,
$a =-5\,m / s ^2$
$F =-0.25\,N$
Standard 11
Physics
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