Two masses as shown in figure are released from heir their positions. Calculate the velocity with which the mass of 5 kg touches the surface if its initial height from the surface is 4 m. Also, show that the gain in KE of the system is equal to loss its P.E. Take g = 10 ms^–2. 

Acceleration of the system,

a =(m₁ − m₂)g/m₁ + m₂

=(5 − 2) × 10/5 + 2

= 4.28m/s²

Velocity with which 5 kg mass touches the ground (v) is obtained from

v² − u² = 2as

v² − 0 = 2 × 4.28 × 4

v = √2 × 4.28 × 4 = 5.85 ms⁻¹.

Initially, both the masses are at rest

∴  Initially KE of the system = 0

Final K.E. of the system = 1/2(m₁ + m₂)v²

= 1/2(5 + 2)(5.85)² = 119.8J

Gain in K.E. = 119.8 − 0 = 119.8J

Initial P.E. of the system = m₁gh₁ + m₂gh₂

= 5 × 10 × 4 + 0 = 200J

Final P.E. of the system = m₁gh₁ + m₂gh₂

= 0 + 2 × 10 × 4 = 80J

Loss in P.E. = 120 − 80 = 120 J.

Thus Gain in K.E. = Loss in P.E.