A .809 kg wooden block is initially at rest on a level surface. A 4.08 g bullet,

Question

A .809 kg wooden block is initially at rest on a level surface. A 4.08 g bullet, traveling horizontally with a velocity of magnitude 378 m/s, is fired into the block. The bullet passes through the block and emerges with its speed reduced to 189 m/s. The block slides a distance of 45.4 cm along the surface from its initial position.
a. What is the coefficient of kinetic friction between the block and surface?
b. What is the decrease in kinetic energy of the bullet?
c. What is the kinetic energy of the block the instant after the bullet passes through it?

Explanation / Answer

a) using momentum conservation,

0.00408 x 378 = 0.00408 x 189 + 0.809 x v

v = 0.953 m/s

v^2 - u^2 = 2as

0 - 0.953^2 = 2 x a x 0.454

a = 1 m/s2

a = (mu)g

mu = 1/9.81 = 0.102 .....Ans

b) decrease in K.E. = 0.00408 x (378^2 - 189^2)/2 = 218.61 J

c) K.E. = 0.809 x 0.953^2/2 = 0.367 J

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