Calculating the Coefficient of Friction

[nvallettejr]

1. A 50 kg Skater pushes off horizontally for a distance of 0.4m with a force of 150N. If she travels 80m before stopping, what is the coefficient of friction?

I am completely lost on this problem, can someone please assist me?

Thanks.

[ApexOfDE]

First, you should find skater's velocity after pushing off. This velocity will be initial one in second path.

In the second path, you should find acceleration using v^2 - v0^2 = 2as, then plug a into Newton's 2nd law eq to find c.f.

[nvallettejr]

I'm not quite sure how to find the velocity after the push off. And would our total distance be 79.6m (80-0.4)?

[Cryphonus]

to find the velocity after the push of;

you have your net Force acting on your skater and you have the mass of the skater; so you can use the newton's law F=m.a to find the acceleration during the push of period; Now you have to find the velocity...

you can use X=1/2at^2 to find it you know the acceleration and the displacement during the push of so you can calculate the t....

[nvallettejr]

was your last post unfinished?

[ApexOfDE]

Can you find skater's in 1st path when you have F, s?

[nvallettejr]

so far i have a=3m/s2 t=7.285s v= 21.855m/s

i am still stuck on how to get the friction coefficient from this information.

[ApexOfDE]

When you have initial speed in 2nd path and its distance, you can find a in this path (a < 0). Now from N's 2nd law, you have a, m => F => c.f

v= 21.855m/s
is this termin speed in 1st path? In 1st path, i have: v^2 = 2as = 2 * 3 * 0.4 = 2.4