Question

Two identical 5 kg blocks are moving with same speed of 2 m/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. The work done by external and internal forces are respectively

Answer 1

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.

Answer 2

Answer:

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

the net work done by the forces is 20 J.