If an object is moving at $7$ $7$ $m s^{- 1}$ $ms^-1$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2/g$ , how far will the object continue to move?

Question

If an object is moving at $7$ $7$ $m s^{- 1}$ $ms^-1$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2/g$ , how far will the object continue to move?

1 Answer

Impact of this question

1585 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-29T12:04:52

The distance traveled is $24.5$ $24.5$ $m$ $m$ .

The mass cancels out and so does the value of 'g', and we can use the calculated acceleration and the fact that the final velocity is 0 to find the distance traveled.

Explanation:

The force causing the deceleration will be the frictional force, given by $F_{frict} = μ F_{norm}$ $F_"frict"=muF_"norm"$ , where the normal force $F_{norm} = m g$ $F_"norm"=mg$ , so $F_{frict} = μ m g$ $F_"frict"=mumg$ .

The acceleration of the object will be given by $a = \frac{F_{frict}}{m} = \frac{μ m g}{m}$ $a=F_"frict"/m=(mumg)/m$

The mass cancels, which is convenient since we have not been told its value: $a = μ g = \frac{2}{g} \times g$ $a=mug=2/g xx g$

The g also cancels, so that the acceleration is just equal to $2$ $2$ $m s^{- 2}$ $ms^-2$ . Since it is in the opposite direction to the initial velocity, we can write this as $- 2$ $-2$ $m s^{- 2}$ $ms^-2$ .

Now we can use $v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$ . The object comes to rest, so the final velocity, $v = 0$ $v=0$ $m s^{- 1}$ $ms^-1$ .

$0^{2} = 7^{2} + (- 2) s$ $0^2 = 7^2 +(-2)s$

Rearranging:

$s = \frac{49}{2} = 24.5$ $s=49/2=24.5$ $m$ $m$

Science

Math

Humanities

... and beyond

If an object is moving at $7$ $7$ $m s^{- 1}$ $ms^-1$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2/g$ , how far will the object continue to move?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If an object is moving at 77 ms−1ms^-1 over a surface with a kinetic friction coefficient of uk=2gu_k=2/g, how far will the object continue to move?

1 Answer

Explanation:

Related questions

Impact of this question

If an object is moving at $7$ $7$ $m s^{- 1}$ $ms^-1$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2/g$ , how far will the object continue to move?