- #1

Callmelucky

- 144

- 30

- Homework Statement
- Body starts to slide down the 60degree incline plane and after 20 meters reaches speed of 4 m/s. What is the coefficient of friction?

- Relevant Equations
- acceleration due to gravity under angle ->Fg=mgsin60, Normal force -> Fn=mgcos60, Ffriction = mi(coefficient of friction)*Fn

So basically I need to find the coefficient of friction given the listed information.

What bothers me is that I am getting two different accelerations for two different approaches. When I calculate acceleration using Fg=mgsin60 I do it this way: Fg=mgsin60 -> ma=mgsin60 ->a=gsin60 -> a=8.66. But when I use formula ##v^2=2as## I get a=0.4.

After I got a using Fg=mgsin60 I use Ftr=Fn*mi -> ma=mgcos60*mi -> ##\frac{a}{gcos60}=mi## -> mi(fr coef)= 1.73. Which makes no sense, since body is moving and coef of friction can't be greater or equal to 1 if body is moving.

And using a I got from ##v^2=2as## -> ##\frac{a}{gcos60}=mi## -> mi= 0.08 and that makes sense, BUT the solution in the textbook is 1.65, so I am really confused. I suppose it's mistake in textbook, but why am I getting two different soltions for a?

So yeah, if you could please explain I would be very grateful,

thank you.

What bothers me is that I am getting two different accelerations for two different approaches. When I calculate acceleration using Fg=mgsin60 I do it this way: Fg=mgsin60 -> ma=mgsin60 ->a=gsin60 -> a=8.66. But when I use formula ##v^2=2as## I get a=0.4.

After I got a using Fg=mgsin60 I use Ftr=Fn*mi -> ma=mgcos60*mi -> ##\frac{a}{gcos60}=mi## -> mi(fr coef)= 1.73. Which makes no sense, since body is moving and coef of friction can't be greater or equal to 1 if body is moving.

And using a I got from ##v^2=2as## -> ##\frac{a}{gcos60}=mi## -> mi= 0.08 and that makes sense, BUT the solution in the textbook is 1.65, so I am really confused. I suppose it's mistake in textbook, but why am I getting two different soltions for a?

So yeah, if you could please explain I would be very grateful,

thank you.