Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A  particle of mass m  is at rest at the origin at time  t=0. It is subjected to a force <img alt="F\left ( t \right )= F_{0}e^{-bt}" src="https://learn.careers360.com/latex-image/?F

A  particle of mass m  is at rest at the origin at time  t=0. It is subjected to a force F\left ( t \right )= F_{0}e^{-bt} in the x direction .Its speed v(t) is depicted by which of the following curves ?

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

F(t)=F_{0}e^{-bt}\; \; \; \; \; (Given)

ma=F_{0}e^{-bt}

a=\frac{F_{0}}{m}e^{-bt}

\frac{dv}{dt}=\frac{F_{0}}{m}e^{-bt}\; or\; dv=\frac{F_{0}}{m}e^{-bt}dt

Integrating both sides, we get

\int_{0}^{v}dv=\int_{0}^{t}\frac{F_{0}}{m}e^{-bt}dt

v=\frac{F_{0}}{m}\left [ \frac{e^{-bt}}{-b} \right ]_{0}^{t}=\frac{F_{0}}{mb}\left [ 1-e^{-bt} \right ]

F(t)=Foe^{-bt}

a=\frac{Fo}{m}e^{-bt}

dv = \frac{Fo}{m}e^{-bt} dt

V = \frac{Fo}{m}\left [ \frac{e^{-bt}}{-b} \right ]_{0}^{t}

= \frac{Fo}{mb}\left [ 1- e^{-bt} \right ]

Posted by

manish painkra

View full answer

Similar Questions

Latest Question