A 3.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is

Question

A 3.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190 m/s. The block slides a distance of 72.0 cm along the surface from its initial position.

a) What is the coefficient of kinetic friction between block and surface?

Express your answer using three significant figures.

B)What is the decrease in kinetic energy of the bullet?

Express your answer to three significant figures and include the appropriate units

Explanation / Answer

by conservation of momentum
Initial p of bullet = p of block + final p of bullet
0.00300 (400) = 0.650 (delta(v) + 0.00300 (190)
1.2 = delta(v) +1.22
delta(v) =0.983 m/s [[this is the v of block since it started at zero]]

a)
Ff = mu Fn = mu m g
and
Ff = m a
and from
Vf^2 = Vi^2 + 2 a d
0 = 0.983^2 + 2 a (0.72)
a = 2.40 m/s/s
so
Ff = 0.650 (2.40) = 1.56 N
then
Ff = mu Fn = mu m g
1.56 = mu (0.650) (9.81)
mu = 0.244
b)
initial energy - final energy = change
1/2 (0.00300) (400)^2 - 1/2 (0.00300) (190)^2 =185.85 J

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