A) \[\sqrt{\frac{mk}{2}}\,{{t}^{{}^{1}/{}_{2}}}\]
B) \[\sqrt{mk}\,{{t}^{{}^{-1}/{}_{2}}}\]
C) \[\sqrt{2mk}\,{{t}^{{}^{-1}/{}_{2}}}\]
D) \[\frac{1}{2}\sqrt{mk}\,{{t}^{{}^{-1}/{}_{2}}}\]
Correct Answer: A
Solution :
As the machine delivers a constant power So F. v = constant = k (watts) \[\Rightarrow \,\,\,m\frac{dv}{dt}.v=k\] \[\Rightarrow \,\,\,\,\int{vdv}=\frac{k}{m}\int{dt}\] \[\Rightarrow \,\,\,\frac{{{v}^{2}}}{2}=\frac{k}{m}t\Rightarrow c=\sqrt{\frac{2k}{m}t}\] Now force on the particle is given by \[F=m\frac{dv}{dt}=m\frac{d}{dt}{{\left( \frac{2kt}{m} \right)}^{\frac{1}{2}}}\] \[=\sqrt{2km}.\,\left( \frac{1}{2}{{t}^{-\frac{1}{2}}} \right)\] \[=\sqrt{\frac{mk}{2}}.{{t}^{-\frac{1}{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
