JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2021
- Hard
- MCQ
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 \(F=F_{0}\left(1-\left(\frac{t-T}{T}\right)^{2}\right)\) Where \(F_{0}\) and \(T\) are constants. The force acts only for the time internal \(2 T\). The velocity \(v\) of the particle after time \(2 {T}\) is -
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{4 {F}_{0} {T}}{3 {M}}\)
Step-by-step Solution
Detailed explanation
\({t}=0, {u}=0\)
As given
\({a}=\frac{{F}_{0}}{{M}}-\frac{{F}_{{o}}}{{MT}^{2}}({t}-{T})^{2}=\frac{{dv}}{{dt}}\)
\(\int_{0}^{v} {d} v=\int_{t=0}^{2 T}\left(\frac{F_{o}}{M}-\frac{F_{0}}{M T^{2}}(t-T)^{2}\right) d t\)
\(V=\left[\frac{F_{0}}{M} t\right]_{0}^{2 T}-\frac{F_{0}}{M T^{2}}\left[\frac{t^{3}}{3}-t^{2} T+T^{2} t\right]_{0}^{2 T}\)
\({V}=\frac{4 {F}_{0} {T}}{3 {M}}\)
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