If a 6 kg object moving at 1 m/s slows down to a halt after moving 3 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Dec 29, 2017

$\mu = \frac{1}{80}$

Explanation:

We know that,
${v}^{2} = {u}^{2} + 2 a s$

$\frac{1}{4} = 1 + 2 \times 3 \times a$

$- \frac{3}{4} = 6 \times a$

$a = - \frac{1}{8}$

By Newton's law,

$F = m a$

$f = 6 \times \frac{1}{8}$

$\mu N = \frac{3}{4}$

$\mu = \frac{3}{4 \times 60}$

$\mu = \frac{1}{80}$