Two identical 5 kg blocks are moving with same speed of `2m//s` towards eachother along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Caluculate work done by (i) external forces (ii) internal forces.

Explanation

Here, `m_(1)=m_(2)=m=5kg, v=2m//s`
As no external force is applied on the system, `F_(ext)=0`
`:. W_(ext.)=vecF_(ext.).vecs=Zero`
According to work energy principle ,
total work done `=` change is KE
`W_(ext.)+W_(i n t.)=0-((1)/(2)mv^(2)+(1)/(2)mv^(2))`
`0+W_(i nt.)=-mv^(2)=-5(2)^(2)=-20J`
`W_(i nt.)=-20J`