Question

The velocity time graph of an object mass 50 g is shown in figure study graph and answer
1)calculate force acting on object in time interval 0-3 seconds
2)calculate the force acting on the object in the time interval 6-10 seconds
3)Is there any time interval in which no force acts on object.Justify

Answer 1

1) acceleration = v-u/t

a= 120-0/3 =40m/s

m= 50/1000 ( as u have to convert g to kg)

m= 1/20

F= ma = 1/20 x 40 = 2N

2) 0x120 / 10-6 = -120/4 = -30m/s^2

1/20 x -30 = -1.5 N

3) velocity of the object is constant and so the acceleration is 0.

Hope it helps....!

Answer 2

Answer:

1. Force acting on the object in the time interval 0-3sec:

  Given initial velocity (u) = 0m/s and final velocity(v) = 120m/s

           acceleration  $a=\frac{v-u}{t}$  

                                  $a=\frac{(120-0)m/s}{3s}$

                                  $a=40 m/s^{-2}$            

        mass of object, $m=50g$

                                   $m=\frac{50kg}{1000}=0.05kg$

     Force acting on the object,

                                     $F=ma$

                                     $F=(0.05kg)(40ms^{-2} )=2kgms^{-2}$

                                     $F=2N$          

2.  Force acting on the object in the time interval 6-10sec:

     Now,   u = 120m/s and v = 0m/s

                 acceleration $a=\frac{v-u}{t}$

                                       $a=\frac{(0-120)m/s}{4s}$

                                        $a= -30m/s^{-2}$

     Force acting on the object,

                                       $F=ma$

                                       $F=(0.05kg)(-30ms^{-2} )=-1.5kgms^{-2}$

                                        $F=-1.5N$    

3.   The time interval when no force acting on the object:

             For the time interval from 3 to 6sec,  the initial and final velocity are same . Therefore, the acceleration is zero and no force acting on the object.