Velocity-time graph for a body of mass 10 kg is shown in figure.  Work-done on the body in first two seconds of the motion is :

  • Option 1)

     12000 J

  • Option 2)

    −12000 J

  • Option 3)

    −45000 J

  • Option 4)

    −9300 J

     

 

Answers (2)

As we discussed in

Net work done by all the forces give the change in kinetic energy -

W=\frac{1}{2}mv^{2}-\frac{1}{2}mv{_{0}}^{2}

W= k_{f}-k_{i}

- wherein

m=mass \: of\: the\: body

v_{0}= initial\: velocity

v= final\: velocity

 

 

 Initial velocity = 50 m/s

Final velocity = 0 

Therefore a=\frac{v-u}{t}=-5 m/s^{2}

Therefore at any time t, velocity is 

v = 50 - 5 t

at t = 0, v = 50 m/s

at t = 2s, v = 40 m/s

Therefore done in first 2 second will be change in kinetic energy

Therefore W=\frac{1}{2}m(v_{f}^{2}-v_{i}^{2})=\frac{1}{2} \times10 \times(1600-2500)

W = - 4500 J

 


Option 1)

 12000 J

This is an incorrect option.

Option 2)

−12000 J

This is an incorrect option.

Option 3)

−45000 J

This is the correct option.

Option 4)

−9300 J

 

This is an incorrect option.

N neha

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