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## Homework Statement

40-g Bullet is fired with a velocity of magnitude v0=700 m/s into a 5kg block of wood. Coefficient of kinetic friction = 0.6.

a) how far will the block move

b) how long after impact does it come to rest

c) percentage of energy lost on impact

(block has no velocity until bullet is completely embedded in block)

## Homework Equations

## The Attempt at a Solution

mass bullet: m = 40g = .04kg

mass block: M = 5kg

velocity bullet = 700 m/s

uk = 0.6

KE of bullet

1/2*0.04*700^2 = 9800J

Conservation of Momentum Work Energy

mv+0 = (M+m)v -uk(m+M)gd=-1/2(m+M)v^2

0.04(700) = 5.04v ukgd=1/2v^2

v = 5.56m/s 0.6(9.81)d=0.5(5.56)^2

d = 2.63

a) 2.63m

b) D=v*t t=d/v

t = 2.63/5.56

t= 0.473s

c) kinetic energy bullet = 9800J

Kinetic energy of block and bullet = 77.9J

%= 77.9/9800 == 0.07%

I am most confused on the percentage of energy lost?