Question 113: Laws of Motion

Question

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 $$F = {F_0}\left[ {1 - {{\left( {{{t - T} \over T}} \right)}^2}} \right]$$Where F0 and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is :

Accepted Answer

At t = 0, u = 0
$$a = {{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{(t - T)^2} = {{dv} \over {dt}}$$
$$\int\limits_0^v {dv = \int\limits_{t = 0}^{2T} {\left( {{{{F_0}} \over M} - {{{F_0}} \over {M{T^2}}}{{(t - T)}^2}} \right)dt} } $$
$$V = \left[ {{{{F_0}} \over M}t} \right]_0^{2T} - {{{F_0}} \over {M{T^2}}}\left[ {{{{t^3}} \over 3} - {t^2}T + {T^2}t} \right]_0^{2T}$$
$$ \Rightarrow $$ $$V = {{4{F_0}T} \over {3M}}$$

Answer

2F0T/M

Answer

F0T/2M

Answer

4F0T/3M

Answer

F0T/3M

Laws of Motion

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