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

Explanation

Here, `m_(1)=5kg, m_(2)=2kg,`
`g=10ms^(-2)`
Acceleration of the system,
`a=((m_(1)-m_(2))g)/(m_(1)+m_(2))=((5-2)xx10)/(5+2)=4.28m//s^(2)`
Velocity with which 5kg mass touches the ground (v) is obtained from
`v^(2)-u^(2)=2as`
`v^(2)-0=2xx4.28xx4`
`v=sqrt(2xx4.28xx4)=5.85ms^(-1)`
Initially, both the masses are at rest
`:.` Inital KE of the system `=0`
Final K.E. of the system `=(1)/(2)(m_(1)+m_(2))v^(2)`
`=(1)/(2)(5+2)(5.85)^(2)=119.8J`
Gain in K.E. `=119.8-0=119.8J`
Initial P.E. of the system `m_(1)gh_(1)+m_(2)gh_(2)`
`=0+2xx10xx4=80J`
Loss in P.E. `=200-80=120J`
Thus , Gain in K.E. `=` Loss in P.E.