Question:
A particle of mass $50 \mathrm{~g}$ 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.
Solution:
Slope of v-t graph gives acceleration,
$\mathrm{t}=25, \mathrm{a}=$ slope of straight line from $\mathrm{t}=0$ to $\mathrm{t}=35=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}^{2}$
$t=45, \quad a=0$
$\left(a s \frac{d v}{d t}=0\right)$
$t=65, \quad a=\frac{-10}{2}=-5 \mathrm{~m} / \mathrm{s}^{2}$
$F(t=25)=0.05 \times 5=0.25 \mathrm{~N}$
$F(t=45)=0.05 \times 0=0 \mathrm{~N}$
$F(t=65)=0.05 \times-5=-0.25 \mathrm{~N}$
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