Questions from Fundamentals of physics 8th editon-Friction

[luthien09]

Does anyone knows how to solve problem 8 on page 130??

Homework Statement

>> In fig. a horizontal force of 100N is to be applied to a 10kg slab that is initially stationary on a frictionless floor, to accelerate the slab. A 10kg block lies on top of the slab;
the coeffitiont of the friction (mu) between the block and the slab is not known,
and the block might slip

(a) considering that possibility, what is the possible range of values for the magnitude
of the slab's acceleration?

(b) what is the possible range for the magnitude a of the block's acceleration?

Homework Equations

F=(mu)*Normal Force
F=ma


3. The Attempt at a Solution



>>I thought like... when the net force of the block is less than the fiction,
the block and the slab move together...
so (a) slab's acc. is ..

F= (M+m)a (when moving together)
a=5m/s^2

F-f(friction)=m(slab's mass)a
100-(mu)*(blockmass)*(g)=ma


answer: 5<a<(100-mu*blockmass*g/m) ?

(b)
block's acceleration
5m/s^2 (when moving together)
f(friction)=ma
a=mu*g

mu*g<a<5??




I'm totally stuck in this problem ...
can anyone help me?
** picture included

 

[Doc Al]

Your solutions look OK to me. It's not clear if they want a numerical range or a range in terms of the unknown μ. I would have answered with a numerical range.

For the slab: As you found, the minimum acceleration will be when the block sticks to it. To get the maximum acceleration, assume there's no friction at all.

For the block, the reasoning is reversed: The maximum acceleration will be when it sticks to the slab; the minimum will be if there were no friction at all.

 

[nlightner]

with the problem for reference **

I would be happy to assist you with this problem. First, let's review the given information. We have a horizontal force of 100N applied to a 10kg slab that is initially stationary on a frictionless floor. On top of the slab, there is a 10kg block with an unknown coefficient of friction (mu) between the block and the slab. We are asked to determine the possible range of values for the accelerations of both the slab and the block.

To solve this problem, we can use Newton's second law, F=ma, where F is the net force acting on the object, m is the mass of the object, and a is the acceleration. We know that the net force on the slab is 100N, and we can assume that the net force on the block is also 100N since the block is not moving initially.

(a) For the slab's acceleration, we can set up the following equation:
F - f = ma
Where F is the applied force, f is the frictional force, and m is the mass of the slab. We can rearrange this equation to solve for a:
a = (F-f)/m
Since we do not know the coefficient of friction, we can use the maximum possible value, which is when the block is just about to slip. This means that the frictional force (f) is equal to the maximum static frictional force, which is given by f = mu*N, where N is the normal force between the block and the slab. The normal force is equal to the weight of the block, which is mg. Therefore, we can rewrite the equation as:
a = (F - mu*mg)/m
Plugging in the values, we get:
a = (100 - mu*10*9.8)/10
Simplifying, we get:
a = 10 - 9.8mu
Since we do not know the value of mu, we can say that the possible range of accelerations for the slab is:
0 < a < 10 m/s^2
This means that the acceleration of the slab can be anything between 0 and 10 m/s^2, depending on the value of mu.

(b) For the block's acceleration, we can use the same equation, but this time we will use the maximum possible value for the coefficient of friction, which is when