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.
So I get that (1/2)MV^2 = MgX*sin 15 + Mg*muk*cos 15*X, but what is V, in this case?
Oh ! Hey you have friction here and are losing energy to heat. You are not allowed to use kinetic energy goes to potential energy if there are frictional losses in the problem. You have to use force equals mass times acceleration here.
call mass of cap = M
then gravity gives a weight, M g = 9.8 M
That 9.8M has a component down the ramp of
9.8 M sin 15 = 2.33 M
and it has normal component on the ramp of
9.8 M cos 15 = 9.47 M
with the friction coef of .35 we have a retarding frictional force down the ramp of
.35*9.47 M = 3.31 M
Both forces are down the ramp so the total force stopping the bottle cap is F = - (2.33+3.31)M - 5.64 M is positive is up the ramp
so get the acceleration
a = F/M = -5.64 M / M
a = - 5.64 m/s^2
Now:
we have initial speed up the ramp, vo = +1.92 m/s
and a = - 5.64 m/s^2
so
v = 1.92 - 5.64 t
when it stops v = 0
0 = 1.92 - 5.64 t
solve for t, the time to stop. Then get distance from
x = 0 + vo t -(1/2)(5.64) t^2
V is 1.92 m/s, the initial velocity.
That looks like an equation I gave you earlier. I believe it is valid because it allows conversion of initial kinetic energy into both heat and gravitational potential. See is you get the same answer that way as by Damon's method.
Bet it goes a lot higher your way :)
OIC - I take that back. I did not see that you had included the work done by friction.
I keep getting 0.315 m, but that's not right. What am I doing wrong?
Well I got .3268
from
t=1.92/5.64
=.3404
and
x = 1.92(.3404) - (5.64)(.5)(.3404)^2
Nope. I made a calculator error. I agree with your .315 meters
Redoing it with calculation typo fixed:
call mass of cap = M
then gravity gives a weight, M g = 9.8 M
That 9.8M has a component down the ramp of
9.8 M sin 15 = 2.53 M [TYPO WAS HERE]
and it has normal component on the ramp of
9.8 M cos 15 = 9.47 M
with the friction coef of .35 we have a retarding frictional force down the ramp of
.35*9.47 M = 3.31 M
Both forces are down the ramp so the total force stopping the bottle cap is F = - (2.53+3.31)M - 5.85 M is positive is up the ramp
so get the acceleration
a = F/M = -5.85 M / M
a = - 5.85 m/s^2
Now:
we have initial speed up the ramp, vo = +1.92 m/s
and a = - 5.64 m/s^2
so
v = 1.92 - 5.85 t
when it stops v = 0
0 = 1.92 - 5.85 t
t = .328 seconds
Then get distance from
x = 0 + vo t -(1/2)(5.85) t^2
x = 1.92 (.328) - 2.93 .328^2
x = .630 - .315
x = .315
and I get the same thing the way Dr WLS did it.
Redoing it with calculation typo fixed:
call mass of cap = M
then gravity gives a weight, M g = 9.8 M
That 9.8M has a component down the ramp of
9.8 M sin 15 = 2.53 M [TYPO WAS HERE]
and it has normal component on the ramp of
9.8 M cos 15 = 9.47 M
with the friction coef of .35 we have a retarding frictional force down the ramp of
.35*9.47 M = 3.31 M
Both forces are down the ramp so the total force stopping the bottle cap is F = - (2.53+3.31)M - 5.85 M is positive is up the ramp
so get the acceleration
a = F/M = -5.85 M / M
a = - 5.85 m/s^2
Now:
we have initial speed up the ramp, vo = +1.92 m/s
and a = - 5.64 m/s^2
so
v = 1.92 - 5.85 t
when it stops v = 0
0 = 1.92 - 5.85 t
t = .328 seconds
Then get distance from
x = 0 + vo t -(1/2)(5.85) t^2
x = 1.92 (.328) - 2.93 .328^2
x = .630 - .315
x = .315
and I get the same thing the way Dr WLS did it.