Problem 6.64 sace trig t onta Foe-t/T is exerted on the object. At t 0, an object of mass m is at
rest at z = 0 on a horizontal, frictionless surface. Starting at t = 0, a horizontal force
Solution
The force acting on the particle is given as Foe-t/T
We know that acceleration = Force / mass
Hence, acceleration = a(t) = (Fo/m)e-t/T
Further, for any motion, we know that dv /dt = a(t)
Rearrganging the terms, we get:
dv = a(t) dt
or dv = (Fo/m)e-t/T dt
We will have to integrate the above relation to get the velocity at any time t
Part A: Hence, V = -(FoT/m)[e-t/T - 1] = (FoT/m)[1 - e-t/T]
Part B:
For the above relation of V, we need to find the value of V for t tending to infinity. For t tending
to infinity, e-t/T will tend to zero.
Hence, Velocity after a very long time = (FoT/m)
NOTE: For questions involving force or acceleration dependent on time, we would need to
employ dv/dt = a(t) to determine the relation for dV which can then be integrated to obtain V at
any point. Plus, similar relation as
ds/dt = v(t) can be used in case displacement is to be determined..
Problem 6.64 sace trig t onta Foe-tT is exerted on the object. At t .pdf
1. Problem 6.64 sace trig t onta Foe-t/T is exerted on the object. At t 0, an object of mass m is at
rest at z = 0 on a horizontal, frictionless surface. Starting at t = 0, a horizontal force
Solution
The force acting on the particle is given as Foe-t/T
We know that acceleration = Force / mass
Hence, acceleration = a(t) = (Fo/m)e-t/T
Further, for any motion, we know that dv /dt = a(t)
Rearrganging the terms, we get:
dv = a(t) dt
or dv = (Fo/m)e-t/T dt
We will have to integrate the above relation to get the velocity at any time t
Part A: Hence, V = -(FoT/m)[e-t/T - 1] = (FoT/m)[1 - e-t/T]
Part B:
For the above relation of V, we need to find the value of V for t tending to infinity. For t tending
to infinity, e-t/T will tend to zero.
Hence, Velocity after a very long time = (FoT/m)
NOTE: For questions involving force or acceleration dependent on time, we would need to
employ dv/dt = a(t) to determine the relation for dV which can then be integrated to obtain V at
any point. Plus, similar relation as
ds/dt = v(t) can be used in case displacement is to be determined.
![Problem 6.64 sace trig t onta Foe-t/T is exerted on the object. At t 0, an object of mass m is at
rest at z = 0 on a horizontal, frictionless surface. Starting at t = 0, a horizontal force
Solution
The force acting on the particle is given as Foe-t/T
We know that acceleration = Force / mass
Hence, acceleration = a(t) = (Fo/m)e-t/T
Further, for any motion, we know that dv /dt = a(t)
Rearrganging the terms, we get:
dv = a(t) dt
or dv = (Fo/m)e-t/T dt
We will have to integrate the above relation to get the velocity at any time t
Part A: Hence, V = -(FoT/m)[e-t/T - 1] = (FoT/m)[1 - e-t/T]
Part B:
For the above relation of V, we need to find the value of V for t tending to infinity. For t tending
to infinity, e-t/T will tend to zero.
Hence, Velocity after a very long time = (FoT/m)
NOTE: For questions involving force or acceleration dependent on time, we would need to
employ dv/dt = a(t) to determine the relation for dV which can then be integrated to obtain V at
any point. Plus, similar relation as
ds/dt = v(t) can be used in case displacement is to be determined.](https://image.slidesharecdn.com/problem6-230704004255-2ccefb8c/85/Problem-6-64-sace-trig-t-onta-Foe-tT-is-exerted-on-the-object-At-t-pdf-1-320.jpg)
