# physics

A 6.25gram bullet traveling at 365ms-1 strikes and enters a 4.50kg crate.The crate slides 0.15m along a wood floor until it comes to a rest.

4. What is the coefficient of dynamic friction between crate and the floor?
5. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?

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1. bullet mass = 0.00625 kg
initial momentum = 0.00625*365
= 2.28 kg m/s

mass after crash = 0.00625+4.50 = 4.506 kg
same momentum so
v = 2.28/4.506 = 0.506 m/s
work done by friction = F d
= F * 0.15 m
so
F = loss of Ke/0.15 = 6.67*loss of Ke
but
the loss of Ke is all of it (1/2)mv^2
=(1/2)(4.506)(.506)^2 = 0.577 Joules
so
F = 6.67 * .577 = 3.85 Newtons
that is mu*4.506*9.81
so
mu = 0.087

5. then Force = change of momentum/time
change of speed of bullet = 364.5 m/s
time = .0110 meters/avergage speed
= .0110m/182 m/s
so force = mass of bullet* 364.5 /(.011/182)

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