Q 1 :

The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the brakes on cars A and B, respectively, then the ratio FA/FB is               [2025]
 

  • 13

     

  • 12

     

  • 32

     

  • 23

     

(4)

According to work - energy theorem,

Work done = change in kinetic energy

W=ΔKE

For car A, WA=(KE)A

FASA=(KE)A                                                 ...(i)

Similarly for car B, FBSB=(KE)B                    ...(ii)

From eqn. (i) and (ii)

(KE)A(KE)B=FASAFBSB100225=FAFB×10001500FAFB=23



Q 2 :

An object moving along horizontal x-direction with kinetic energy 10 J is displaced through x=(3i^)m, by the force F=(-2i^+3j^)N. The kinetic energy of the object at the end of the displacement x is:                          [2024]

  • 10 J

     

  • 16 J

     

  • 4 J

     

  • 6 J

     

(3)

Given, F=(-2i^+3j^)N or x=(3i^)m

Initial Kinetic energy, (Ki)=10J

         W=F·x=(-2i^+3j^)·(3i^)=-6J

Using work-energy theorem, W=ΔK.E.

-6=Kf-Ki or -6=Kf-10; Kf=4J

 



Q 3 :

Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 ms-1. Take g constant with a value 10 ms-2. The work done by the (i) gravitational force and the (ii) resistive force of air is              [2017]
 

  • (i) 1.25 J                 (ii) - 8.25 J

     

  • (i) 100 J                 (ii) 8.75 J

     

  • (i) 10J                  (ii) - 8.75 J

     

  • (i) - 10J                 (ii) - 8.25 J

     

(3)

Here, m=1 g=10-3kg,

h=1km=1000m, v=50m s-1, g=10m s-2

(i) The work done by the gravitational force 

           =mgh=10-3×10×1000=10J

(ii) The total work done by gravitational force and the resistive force of air is equal to change in kinetic energy of rain drop.

   Wg+Wr=12mv2-0

        10+Wr=12×10-3×50×50 or Wr=-8.75J



Q 4 :

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10-4J by the end of the second revolution after the beginning of the motion?           [2016]
 

  • 0.18m/s2

     

  • 0.2m/s2

     

  • 0.1m/s2

     

  • 0.15m/s2

     

(3)

Here, m=10 g=10-2kg, R=6.4cm=6.4×10-2m,

Kf=8×10-4J, Ki=0, at= ?

Using work energy theorem,

Work done by all the forces = Change in KE

Wtangential force+Wcentripetal force=Kf-Ki

at=Kf4πRm=8×10-44×227×6.4×10-2×10-2=0.0990.1ms-2