If a 6 kg object moving at 12 m/s slows down to a halt after moving 9 m, what is the friction coefficient of the surface that the object was moving over?

Jun 24, 2017

${\mu}_{k} = 0.82$

Explanation:

The equation for friction in terms of friction coefficient is:
$F = {\mu}_{k} N$
Where ${\mu}_{k}$ is the kinetic friction coefficient.

To solve this we therefore need to find the magnitude of the friction force and the normal reaction force.

1. Find the magnitude of the normal reaction
We assume that the only vertical forces acting on the object are weight and normal reaction. Therefore the normal reaction equals the weight…
N = w = mg = 6 × 9.8 = 58.8 N

2. Find the magnitude of the friction force
We know the initial velocity of the object and over which distance it decelerates to a halt, so we also know that the final velocity is zero as it came to a halt. If we assume that the deceleration due to the friction is constant then we can use an equation of constant acceleration to find the deceleration and hence the friction force.

$s = 9$ m
$u = 12$ m s⁻¹
$v = 0$
a=?
t =×
Use equation ${v}^{2} = {u}^{2} + 2 a s$
=> a = (v^2 - u^2)/(2s) = (0 - 144)/(2×9) = -8.0 m s⁻¹

Now use Newton's second law of motion to solve for the force:
F = ma = 6.0 × (-) 8.0 = (-)48 N
I have put the negatives in brackets as we are solving for the magnitude and the negative tells us about the direction of the force and acceleration (i.e. the force is decelerating the object).

**3. Lastly calculate the coefficient F = mu_kN => mu_k = F/N = 48 / 58.8 = 0.8163 ≈ 0.82#