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2 days ago

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A 1.7-kg block of wood rests on a rough surface. A 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself.

The bullet-block system slides forward 2.4 m before coming to rest.

1 answer:

ValentinkaMS [3.3K]2 days ago

7 0

What is the coefficient of kinetic friction between the block and the surface? Provide your answer using two significant figures.

Response:

0.39

To elaborate:

The provided data includes

m1=1.7 Kg

m2=0.011 Kg

v2=670 m/s

d=2.4 m

The relationship m2v2=(m1+m2)v applies here hence $v=\frac {m2v2}{m1+m2}$ also $a=\frac {v^{2}}{2d}$

Friction's deceleration can be calculated using

$F=\mu_k N=\mu_k W=\mu_k (m1+m2)g$

$F=(m1+m2)a=\mu_k (m1+m2)g$ thus $a=\mu_k g$

$a=\frac {v^{2}}{2d}=\mu_k g$

$\mu_k=\frac {1}{2dg}(\frac {m2v}{m1+m2})^{2}=\frac {1}{2*2.4*9.81}\times (\frac {0.011*670}{1.7+0.011})^{2}$

$\mu_k=0.39$

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