# A 6.0 kg block is attached to a spring with k = 120kg/s?

A 6.0 kg block is attached to a spring with k = 120kg/s, and slides on a

floor with a coefficient of kinetic friction mk = 0.25. The block begins at the equilibrium

position. It is struck by a bullet with a mass of 15 grams, moving at 650 m/s and coming

from the right. The bullet becomes embedded in the block.

How far to the left does the block move?

### 1 Answer

- JullyWumLv 79 years agoFavorite Answer
Assuming that k = 120 kg/s² = 120 N/m

Finding initial vel.(V) of block after impact by cons.of momentum ..

Initial mom. = 0.015kg x 650m/s = 9.75kg.m/s .. (for bullet)

Final mom. = 6.015kg x V .. (joint mass)

V = 9.75 / 6.015 = 1.62 m/s

Suppose block comes to rest after a distance d (metres) ..

• Work done against friction (transferring energy from block) =

Friction(N) x d = 0.25(mg)d = (0.25*6.0*9.8)d = 14.70*d J

• Energy transferred to spring from block = ½ kd² = ½*120*d² = 60.0d²

Applying cons.of energy as block comes to rest ..

Initial KE of block = Energy transferred by friction + energy transferred to spring

½ 6.0*(1.62)² = 14.70d+ 60d²

60d² + 14.70d - 7.87 = 0 ( we have a quadratic to solve for d ♦)

►d = 0.26m

♦ For quadratic solution, d = {-b + √(b² - 4ac)} / 2a

.. a= 60, b= 14.7, c= -7.87

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