Question

If an object is moving at `4 ms^-1` over a surface with a kinetic friction coefficient of `mu_k=16 /g`, how far will the object continue to move?

Answer 1

The object will decelerate (accelerate in a direction opposite to its velocity) and stop after `1/2 m` due to the frictional force acting on it.

Explanation:

If there were no friction, the object would keep moving together. Since there is friction, there is a net unbalanced force acting that will cause the object to slow and stop.

Summarising what we know and what we need to know:

`mu_k=16/g` - coefficient of kinetic friction
`u = 4ms^-1` - initial velocity
`v = 0ms^-1` - final velocity
`d = ? m` - distance to stop

We will need to calculate the acceleration of the box, `a (ms^2)`, to calculate the distance. We use Newton's Second Law :

`a=F/m`

Now the force will just be the frictional force, which is the frictional coefficient times the weight force of the object, `mg`:

`F_f = F_(weight)*mu_k = mg*16/g = 16 m` (g cancels)

Substituting into Newton's Second Law:

`a = F_f/m = (16m)/m = 16 ms^-2`

Now we have `v, u, a` and `d`, so we will use:

`v^2 = u^2 + 2ad`

Rearranging to make `d` the subject:

`d = (v^2 - u^2)/2a = (0^2 - 4^2)/(2*16) = 1/2 m`

So the object will stop after `1/2m` or `0.5m`.

(There is an issue with this question: a coefficient of friction should be 'dimensionless' (have no units), but `g` has the units of `ms^-2`. It's on the bottom line so the units over all would be `m^-1s^2`. I think what the person who set the question was trying to do was have `g` cancel out in the calculation, as you saw above, but it's poor physics. A coefficient of friction is just a number.)

Answer 2

I found `0.5m`

Explanation:

From Newton's Second Law, `SigmavecF=mveca`, you can see that the object will receive an acceleration opposite to the direction of movement (deceleration) caused by friction `f_k=mu_k*N` (where in this case the normal reaction `N` is equal to the weight and given as: `N=mg`) and given as:

`-f_k=ma`

so that:
`a=-f_k/m=-(mu_k*N)/m=-(mu_k*mg)/m=-mu_k*g=`
`=-16/g*g=-16m/s^2`

with this information we can use the kinematic relationship:

`v_f^2=v_i^2+2ad`
where the final velocity will be zero:
`0^2=4^2-2*16d`
and
`d=0.5m`