A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static

step 1 Okay, so for this problem, we've got these two blocks. We've got a big slab that weighs, that ha

A 40 kg slab rests on a frictionless floor. A 10 kg block...

A $40 \mathrm{~kg}$ slab rests on a frictionless floor. A $10 \mathrm{~kg}$ block rests on top of the slab. The static coefficient of friction between the block and the slab is $0.60$ while the kinetic coefficient of friction is $0.40$. The $10 \mathrm{~kg}$ block is acted upon by a horizontal force of $100 \mathrm{~N}$. If $g=9.8 \mathrm{~ms}^{-2}$, the resulting acceleration of the slab will be (a) $1.47 \mathrm{~ms}^{-2}$ (b) $1.69 \mathrm{~ms}^{-2}$ (c) $9.8 \mathrm{~ms}^{-2}$ (d) $0.98 \mathrm{~ms}^{-2}$

A $40 \mathrm{~kg}$ slab rests on a frictionless floor. A $10 \mathrm{~kg}$ block rests on top of the slab. The static coefficient of friction between the block and the slab is $0.60$ while the kinetic coefficient of friction is $0.40$. The $10 \mathrm{~kg}$ block is acted upon by a horizontal force of $100 \mathrm{~N}$. If $g=9.8 \mathrm{~ms}^{-2}$, the resulting acceleration of the slab will be
(a) $1.47 \mathrm{~ms}^{-2}$
(b) $1.69 \mathrm{~ms}^{-2}$
(c) $9.8 \mathrm{~ms}^{-2}$
(d) $0.98 \mathrm{~ms}^{-2}$

Key Concepts

Relative Motion and Interaction Forces

In systems involving multiple bodies in contact, it's important to analyze the relative motion between them. This concept involves determining if the force applied to one object is sufficient to overcome the static friction with another, leading to different accelerations, or if the bodies move together as a single system with a shared acceleration.

Free-Body Diagram and Force Analysis

Using free-body diagrams to represent the forces acting on each body allows for a systematic analysis of the problem. By breaking down the forces—including the external applied force and frictional forces—on both the slab and the block, one can apply Newton's laws effectively to determine the resulting accelerations.

Static and Kinetic Friction

Understanding the difference between static and kinetic friction is crucial in problems involving contacting surfaces. Static friction acts to prevent relative motion up to a certain maximum value, while kinetic friction comes into play once sliding starts, generally with a lower coefficient. These concepts determine whether the block moves together with the slab or slips relative to it.

Newton’s Second Law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is essential when analyzing forces on the slab and block, as it provides the framework for calculating their motions based on the applied forces and masses.

Transcript

00:01 Okay, so for this problem, we've got these two blocks.

00:04 We've got a big slab that weighs, that has a mass of 40 kilograms.

00:09 We've got a smaller cube box that has a mass of 10 kilograms.

00:14 And we are going to be pulling on the smaller mass with a force of 100 newtons.

00:21 And we're told that the static coefficient is, the coefficient of static friction is 0 .6.

00:27 And the coefficient of kinetic friction is 0 .4.

00:29 And this is between the two blocks.

00:32 Then we need to figure out, a, what the acceleration of the little box is, and b, what the acceleration of the larger boxes.

00:40 So first of all, we need to figure out, is this going to move together as one big thing, or is the smaller mass going to slide along the larger mass? to do that, we're going to figure out the maximum force of friction.

00:58 So we're just going to look at the vertical forces acting on the small mass.

01:04 So the net force acting on the small mass is going to be zero.

01:08 The downward force, there's going to be a downward force of gravity, f sub g, and there's going to be an upward force of the normal force, f sub n.

01:19 And so we've got f sub g down and f sub n up.

01:25 And so this is just mass times gravity.

01:28 And so we can say that the normal force acting on this is equal to mass times gravity.

01:35 So small m times g.

01:38 That means that the friction force between these two is going to be mu times the mass times gravity.

01:47 So we can figure out the maximum friction force.

01:50 So f friction max happens with the maximum static coefficient of friction.

01:57 So let's see what that is.

02:00 So, mu -static times mass times gravity.

02:04 So this is that normal force.

02:05 And we're going to get 0 .6 times 10 times 9 .8 or 58 .8 newtons.

02:14 Now this is less than the force that we're applying for 100 newtons...