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  • A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation                  &

A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation                       \mathrm{F}=\mathrm{F}_{0}\left[1-\left(\frac{\mathrm{t}-\mathrm{T}}{\mathrm{T}}\right)^{2}\right]
Where \mathrm{F}_{0}$ and $\mathrm{T}are constants. The force acts only for the time interval 2 \mathrm{~T}. The velocity \mathrm{v} of the particle after time 2 \mathrm{~T} is :
 
Option: 1 2 \mathrm{~F}_{0} \mathrm{~T} / \mathrm{M}
Option: 2 F_{0} T / 2 M
 
Option: 3 4 \mathrm{~F}_{0} \mathrm{~T} / 3 \mathrm{M}
 
Option: 4 F_{0} T / 3 M

Answers (1)

best_answer

\begin{aligned} & F=F_{0}\left[1-\left(\frac{t-T}{T}\right)^{2}\right] \\ \end{aligned}

M\frac{dv}{dt}=F_{0}\left [ 1-\frac{\left ( t-T \right )^{2}}{T^{2}} \right ]

dv=\left ( \frac{F_{0}}{M} \right )\left [ 1-\left ( \frac{t_{2-2Tt+T^{2}}}{T^{2}} \right ) \right ] dt

\int_{0}^{V}dv=\frac{F_{0}}{M}\left [ t-\frac{t^{3}}{T^{2}}+\frac{2T}{T^{2}}\left ( \frac{t^{2}}{2} \right )-t \right ]_{0}^{2T}

\begin{aligned} &V=\frac{F_{0}}{M}\left[2 T-\frac{8 T^{3}}{T^{2}}+\frac{2 T\left(4 T^{2}\right)}{T^{2}(2)}-2 T-0\right] \\ &V_{0}=\frac{F_{0}}{M}[-4 T]=\frac{-4 F_{0} T}{3 M} \end{aligned}

The correct option is (3)

 

Posted by

vishal kumar

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