Question

A man of mass 60kg sitting on ice pushes a block of mass 12kg on ice horizontally with a speed 5m/s.The coefficient of friction between the man and ice and between block and ice is 0.2 . The distance between man and block when come to rest

Answer 1

Answer:

The distance between man and block when come to rest is 6.5 m

Explanation:

According to the problem the mass of the man , M is 60 kg

and mass of the block, m is 12 kg

The velocity of pushing the block ,v1 = 5 m/s

The coefficient friction μ= 0.2

Let the velocity of the man is v2 m/s

Now, from the momentum conservation law

f= 0

Mv1+ mv2= 0

60 u + 12 x (-5) =0

=> v1= 1

Therefore ,

v^2- u^2=  2as

=>v^2- 0 = 2 x  μg x s

=> s = v^2/2μg

let d1 be the distance traveled by the man and d2 is the distance traveled by the block

d1= v^2/2μg = 1^2/2x 0.2 x 10 = 1/4

d2= v^2/2μg = 5^2/ 2 x 0.2 x 10 = 25/4

The total distance between them ,d = d1+d2= 1/4+ 25/4= 6.5 m