1.0kg physics book on slope, and Newton's Third Law?

[Jthunder]

Homework Statement

A 1.0 kg physics book is on a 20 degree slope. It is connected by a string to a 500 g coffee cup dangling at the bottom side of the incline. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are us = .50 and uk = .20.

a. How far does the book slide?
b. At the highest point, does the book stick to the slope, or does it slide back down?

Homework Equations

F=m*a
vf^2 = vi^2 + 2as

The Attempt at a Solution

So I found four forces: Tension of the rope pulled by the weight, fk (friction force), force of gravity at a 20 degree angle, and normal force.
n = mgcosx = (9.8)(1)cos20 N...fk = uk*n = (.2)n = 1.842 N
T = mass of cup * 9.8 = 4.9 N
Fg = mgsinx = (1)(9.8)sin20 = 3.35 N

...so I get (Fnet)x = ma = -4.9N - 1.842N - 3.35N = -10.095
...so... a = (-10.095)/(1) = -10.095 m/s^2

But when I put this value of acceleration into vf^2 = vi^2 + 2as, or 0=9 + 2(-10.095)s, I get .446m.
The correct answer is .67m. What did I do wrong? I'm pretty sure it's around the (Fnet)x part.


Thanks for any help.

 

[tiny-tim]

Welcome to PF!

Hi Jthunder! Welcome to PF!

A 1.0 kg physics book is on a 20 degree slope. It is connected by a string to a 500 g coffee cup dangling at the bottom side of the incline. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are us = .50 and uk = .20.
…
T = mass of cup * 9.8 = 4.9 N

No, the cup is accelerating, so T will be different.

Either do two F = ma equations (one for the book, one for the cup), and eliminate T, or just treat the book and the cup as a single body, as if it was one-dimensional, and calculate the external forces on it, along that dimension.