A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time be x = 0, and predict its position at t = -5 s, 25s, 100s.

Answers (1)

The acceleration of force is given by :

                                          a = fracFm

                                          a = frac- 80.4 =- 20 m/s^2

At  t = - 5 s  : 

                                There is no force acting so accleration is zero and   u =  10 m/s

                                         s = ut + frac12at^2

or                                            = 10(-5) + frac12.0.t^2

or                                            = -  50 m

 

At  t   =   25 s   :

                             Acceleration is  -  20  m/s2    and    u  =  10  m/s

                                         s = ut + frac12at^2

or                                           = 10(25) + frac12(-20)25^2

or                                           = - 6000 m

 

At   t  =  100 s

               We have acceleration for first 30 sec, and then it will move with constant speed.

                So for    0  <  t  <  30  :

                                           s = ut + frac12at^2

 or                                            = 10(30) + frac12(-20)30^2

or                                             = - 8700 m

Now for t  >  30 s  :

                                         We need to calculate velocity at t  = 30 sec which will be used as the initial velocity for    30  <  t  <  100.

                                                 v = u + at

or                                                   = 10 + (-20)30 = - 590 m/s

Now                                         s = vt + frac12at^2

or                                                  = (-590)70 + frac12.0.t^2

or                                                  = - 41300 m

Hence total displacement is :       -  8700   +  (- 41300 )  =   -  50000 m. 

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