How Does Friction Affect the Movement of a Book Up a Sloped Surface?

[vorcil]

1. The 1.0 kg physics book in figure is connected by a string to a 500 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are Frictionstatic=0.5 and FrictionKinetic 0.2.

question 1, a) find how far it goes up the slope
question 2, b) after stopping does it stay at rest or go back down the slope



2. The kinematic equations..
picture of the situation http://www.webassign.net/knight/p8-38.gif.



3.
Fc=force of cup Fb=force of book

Fcy = .5*9.8=4.9N (force of the cup due to gravity)

Fbx = N*sin(angle), (using vectors)
N=1kg*9.8ms^-1=9.8N
9.8*sin20=-8.94N (negative due to the direction downwards)

Friction force opposing motion, N*Fkinetic
9.8N * 0.2 = 1.96N(I'm assuming i have to ignore the static friction because the book is moving, this is correct?)

I add the friction force, The cup force and the force of the book together to get the total force of the book down the slope, which equals 15.8N

I know i then use the formula Vf^2=vi^2+2a*x to solve for x, but i don't know how to get the acceleration of the book up the slope!

Do i take the total force down the slope, 15.8N
use the formula f=ma to get the acceleration down the slope?
what mass do i use? the mass of the book or the mass of the book + mass of the cup to solve for this acceleration?
15.8/(massofBook+cup) = deceleration of the book as it is being pushed up

-

help please XD
This is for the mastering physics knight 2e,
I got the awnser to a) but only by capturing the SWF and hacking it to see the awnser
it was 0.669M (the book is pushed up the slope)
the second question b was a no brainer that dosen't require hacking lol.

 

[LowlyPion]

Friction force opposing motion, N*Fkinetic
9.8N * 0.2 = 1.96N(I'm assuming i have to ignore the static friction because the book is moving, this is correct?)

As long as it is kinetic, ignore the static friction coefficient.

Of course you can calculate right away whether it will remain at rest after coming to rest using the static coefficient. But other than that it's not needed.

As to hacking the answer, ... why? It is straight forward enough. But unfortunately the way you are approaching it is likely making it difficult for you.

The forces are easy peasy:
1/2*g ... the cup
1*g*sin20 ... the book from gravity
μ*1*g*cos20 ... from friction

That yields:

F = m*a = .5*g +sin20*g + (.2)*cos20*g

With total mass at 1.5 kg then

a = (.5 + sin20 + cos20)*g/1.5

 

[nlightner]

As a scientist, it is important to approach problems and questions in a systematic and logical manner. Let's break down the given information and try to solve for the unknowns step by step.

First, let's define our variables:
m1 = mass of book (1.0 kg)
m2 = mass of coffee cup (0.5 kg)
u1 = initial velocity of book (3.0 m/s)
u2 = initial velocity of coffee cup (0 m/s)
a1 = acceleration of book
a2 = acceleration of coffee cup
f1 = coefficient of kinetic friction between book and slope (0.2)
f2 = coefficient of static friction between book and slope (0.5)
g = acceleration due to gravity (9.8 m/s^2)
θ = angle of slope (20 degrees)

Now, let's tackle question 1a: How far does the book go up the slope?

We can use the kinematic equation Vf^2=Vi^2+2ax to solve for x, the distance traveled. Since the book starts from rest (u1 = 0 m/s) and ends with a final velocity of 0 m/s (Vf1 = 0 m/s), the equation becomes:
0 = (3.0 m/s)^2 + 2a1x

Solving for x, we get x = 0.45 m. Therefore, the book travels 0.45 meters up the slope.

Now, for question 1b: After stopping, does the book stay at rest or go back down the slope?

To answer this question, we need to consider the forces acting on the book. The only force acting in the direction of motion is the component of the book's weight parallel to the slope, which is Fbx = m1gsinθ. The forces opposing motion are the force of kinetic friction (Ff1 = f1m1gcosθ) and the normal force (N = m1gcosθ).

Using Newton's second law (ΣF = ma), we can set up the following equation:
Fbx - Ff1 - N = m1a1

Substituting in our known values, we get:
m1gsinθ - f1m1gcosθ - m1gcosθ = m1a1

Simplifying, we get:
a1 = g(sinθ - f1cosθ