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.

 

[kerimek]

As a scientist, it is important to carefully analyze and evaluate the problem and solution in order to determine any potential errors. In this case, it seems that there may be a mistake in the calculation of the net force in the x-direction. It appears that the friction force (fk) may have been incorrectly calculated as 1.842 N instead of 0.3684 N (using the given coefficient of friction uk = 0.20). This could result in a larger negative net force and a smaller acceleration, which would explain the discrepancy in the calculated distance. It is also important to double check all units and conversions to ensure accuracy in the calculations. Additionally, it may be helpful to break down the forces and accelerations in the x and y directions separately to avoid any confusion or errors. Overall, it is important to carefully check and review all steps in the problem solving process to ensure accurate and precise results.