Question

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance.

Can someone please show me the steps to solving this one?

Answer 1

assume the bottle cap mass is m, it will not matter what m is.

There are only two forces on the mass in the direction down the incline, that due to the weight and that due to friction.

the amount due to the weight is:
m g sin 15 down the ramp parallel with the ramp

the amount due to friction is also down the ramp while the bottle cap is moving up

m g (.35) cos 15

so the total force down the ramp is:
m g ( sin 15 + .35)

that force is the mass times the acceleration down the ramp
m g (sin 15 + 3.5) = m ( acceleration down the ramp)

notice that we can divide both sides by m so the mass did not matter

so
a = 9.8 ( sin 15 + .35) DOWN the ramp
FIND that number and use it for a

if x is distance UP the ramp then
x = xo + vo t - .5 a t^2
v = vo - a t

vo = 1.92
vf = .95 = 1.92 - a t

solve that for t, the time during which it slows down
then use that t in

x = 0 + 1.92 t - .5 a t^2

Answer 2

I come up with 0.233 m as my answer, but that's not right. What am I doing wrong?

Answer 3

looks pretty good to me

Answer 4

I have no problem with your answer.

Answer 5

Hey wait a minute, I called cos 15 one. It should be .966
That will change the answer a little.

Answer 6

So wait...do I use cos 15 to find the acceleration?

Answer 7

Yes

Answer 8

But I thought acceleration was a = 9.8 ( sin 15 + .35)...