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# If a 2 kg object moving at 1/2 m/s slows to a halt after moving 5 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Mar 13, 2018

${\mu}_{k} = 0.002548$

#### Explanation:

Well, the force of friction ${F}_{f}$ is defined as the following...

${F}_{f} = {\mu}_{k} N$

And the normal is given by...

$N = m g$

$N = \left(2\right) \left(9.81\right)$

$N = 19.62 N$

So now, we need to calculate an acceleration to get the force of friction. We are assuming in this question that ${F}_{\mathrm{dr} i v e} = {F}_{f}$ because all the energy that the mass possess is lost to friction.

${v}^{2} = {u}^{2} - 2 a s$

${0}^{2} = {0.5}^{2} - 2 \left(5\right) a$

$0 = 0.25 - 10 a$

$a = \frac{- 0.25}{10}$

$a = - 0.025 m {s}^{-} 2$

${F}_{f} = m a$

${F}_{f} = \left(1\right) \left\mid - 0.025 \right\mid$

${F}_{f} = 0.025 N$

${\mu}_{k} = {F}_{f} / N$

${\mu}_{k} = \frac{0.025}{9.81}$

${\mu}_{k} = 0.002548$