# If a 2 kg object moving at 6 ms^-1 slows down to a halt after moving 10 m, what is the friction coefficient of the surface that the object was moving over?

Apr 28, 2016

The object's initial kinetic energy will be used as work overcoming the frictional force. The calculation below shows that the coefficient of friction is approximately 5.4.

#### Explanation:

The object's initial kinetic energy is given by:

${E}_{k} = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times 2 \times {6}^{2} = 36$ $J$

This energy will be dissipated doing work against the frictional force acting on the object.

$W = F d$

Rearranging:

$F = \frac{W}{d} = \frac{36 J}{10 m} = 3.6$ $N$

This is the magnitude of the frictional force, which is given by:

${F}_{\text{frict}} = \mu {F}_{N}$, where ${F}_{N}$ is the normal force, which in turn is the weight force of the object:

${F}_{N} = m g = 2 \cdot 9.8 = 19.6$ $N$

To find the frictional force, then:

$\mu = \frac{{F}_{N}}{{F}_{\text{frict}}} = \frac{19.6}{3.6} \approx 5.4$

(coefficients of friction do not have units)

(this is an improbable value for a coefficient of friction - they are typically less than 1, but occasionally as high as 2. Teachers and textbooks do sometimes write questions with these improbable values, though)