If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ over a surface with a kinetic friction coefficient of $μ_{k} = \frac{16}{g}$ $mu_k=16 /g$ , how far will the object continue to move?

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If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ over a surface with a kinetic friction coefficient of $μ_{k} = \frac{16}{g}$ $mu_k=16 /g$ , how far will the object continue to move?

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Answer 1 · 2016-01-17T23:13:41

The object will decelerate (accelerate in a direction opposite to its velocity) and stop after $\frac{1}{2} m$ $1/2 m$ 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:

$μ_{k} = \frac{16}{g}$ $mu_k=16/g$ - coefficient of kinetic friction
$u = 4 m s^{- 1}$ $u = 4ms^-1$ - initial velocity
$v = 0 m s^{- 1}$ $v = 0ms^-1$ - final velocity
$d = ? m$ $d = ? m$ - distance to stop

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

$a = \frac{F}{m}$ $a=F/m$

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

$F_{f} = F_{w e i g h t} \cdot μ_{k} = m g \cdot \frac{16}{g} = 16 m$ $F_f = F_(weight)*mu_k = mg*16/g = 16 m$ (g cancels)

Substituting into Newton's Second Law:

$a = \frac{F_{f}}{m} = \frac{16 m}{m} = 16 m s^{- 2}$ $a = F_f/m = (16m)/m = 16 ms^-2$

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

$v^{2} = u^{2} + 2 a d$ $v^2 = u^2 + 2ad$

Rearranging to make $d$ $d$ the subject:

$d = \frac{v^{2} - u^{2}}{2} a = \frac{0^{2} - 4^{2}}{2 \cdot 16} = \frac{1}{2} m$ $d = (v^2 - u^2)/2a = (0^2 - 4^2)/(2*16) = 1/2 m$

So the object will stop after $\frac{1}{2} m$ $1/2m$ or $0.5 m$ $0.5m$ .

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

Answer 2 · 2016-01-17T23:19:17

I found $0.5 m$ $0.5m$

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
$d=0.5m$

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If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ over a surface with a kinetic friction coefficient of $μ_{k} = \frac{16}{g}$ $mu_k=16 /g$ , how far will the object continue to move?

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Explanation:

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If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ over a surface with a kinetic friction coefficient of $μ_{k} = \frac{16}{g}$ $mu_k=16 /g$ , how far will the object continue to move?