Two identical 5kg block are moving with same speed of 2m/s towards each other along friction less horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Calculate work done by internal forces.

Open in App

Solution

Initial kinetic energy of the system of two blocks is,
KE_i = ½ (5)(2²) + ½ (5)(2²) = 20 J

After the collision the Ek of the system is, KE_f = 0

Change in KE in magnitude is, |ΔKE| = |KE_f - KE_i| = 20 J

From work energy principle, |W| = |ΔKE| = 20 J

Thus, the net work done by the forces is 20 J.

Suggest Corrections

Similar questions

Q. Two identical 5kg blocks are moving with same speed of

2 m s^{- 1}

towards each other along a frictionless horizontal surface. The two blocks collide, stick together, and come to rest. Consider the two blocks as a system. The work done by external and internal forces are respectively,

Q. Two identical 5 kg blocks are moving with same speed of

2 m / s

towards each other along a friction-less horizontal surface. The two blocks collide, stick together and come to rest. Consider two blocks as a system, the work done on the system by the external forces will be:

Two identical blocks each of mass 10 kg are moving with a speed of 4m/s towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system, work done by external force is

Q. Two identical

4

kg blocks are moving with same velocity

1

/

s towards each other along a smooth horizontal surface, when they collide they stick together and comes to rest. The work done by external and internal forces will be respectively.

Q. Two identical balls each of mass 2 kg. are moving towards each other with the same speed of

5 m / s

. They collide with each other, stick together and come to rest what is the work done by the internal forces-

Join BYJU'S Learning Program

Two identical 5kg block are moving with same speed of 2m/s towards each other along friction less horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Calculate work done by internal forces.

Initial kinetic energy of the system of two blocks is, KEi = ½ (5)(22) + ½ (5)(22) = 20 J After the collision the Ek of the system is, KEf = 0 Change in KE in magnitude is, |ΔKE| = |KEf - KEi| = 20 J From work energy principle, |W| = |ΔKE| = 20 J Thus, the net work done by the forces is 20 J.