Question

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

Answer 1

`0.612` to 3dp.

Explanation:

We assume the kinetic friction coefficient is constant throughout the motion. To calculate the force exerted by friction we need to calculate the acceleration of the object.

`v^2 = u^2 + 2aDeltax`

`therefore a = (v^2-u^2)/(2Deltax)`

`a = (0-6^2)/(2*3) = -6ms^-2`

Using Newton's second law, the force required to induce `-6ms^-2` of acceleration on a `3kg` object is:

`F = ma`

`F = 3*(-6)N = -18N`

Remember that acceleration and force are vectors so the negative sign just denotes it is pointing opposite to the direction we are ascribing as positive.

We now know that the magnitude of the kinetic frictional force is `18N`

`vec(f)_k = mu_kvec(n)`

`mu_k = (|vec(f)_k|)/(|vec(n)|) = 18/(m*g) = 18/(3*9.81) = 0.612` to 3dp.