Question

A 20 g bullet is fired horizontally with a speed of 600 m/s into a 7 kg block sitting on a table top ; the bullet (b) lodges in the block (b). If the coefficient of kinetic friction between the block and the table top is 0.4, then what is the distance the block will slide ?

Answer 1

As we have learned

Coefficient of Friction(µ) -

Ratio of static friction to the normal = F/R

After coliision , common velocity

v = 600 * 20

-------- = 17.14 m/s

700

\mu = 0.6

Decelleartion a =

ųg = 0.6 * 9.8 = 5.88 m/s^2

Distance travelled S =

= u^2 ( 17.14 )^2

---- = --------- = 24.98 m

2a 2 * 5.88

APPROX. 25 m

I HOPE IT WILL HELP YOU .......