A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2. After the collision the bullet becomes embedded in the block. How much work is being dne by bullet?

Work done by the bullet is 612.26 J.

Explanation:

mass of bullet, m = 0.5 kg

initial velocity of bullet, u = 50 m/s

coefficient of friction = 0.2

mass of block, M = 3 kg

let the final speed of the bullet block system is v.

use conservation of momentum

Momentum of bullet + momentum of block = momentum of bullet block system

0.5 x 50 + 3 x 0 = (3 + 0.5) v

v = 7.14 m/s

let the stopping distance is

The work done is given by change in kinetic energy of bullet

initial kinetic energy of bullet, K =  0.5 x 0.5 x 50 x 50 = 625 J

Final kinetic energy of bullet, K' = 0.5 x 0.5 x 7.14 x 7.14 = 12.74 J

So, the work done by the bullet

W = 625 - 12.74 = 612.26 J  

multiply 2kg by latent heat of vap

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2