A block of mass m=2.00 kg rests on the left edge of a block...

A block of mass $m=2.00 \mathrm{~kg}$ rests on the left edge of a block of larger mass $M=8.00 \mathrm{~kg} .$ The coefficient of kinetic friction between the two blocks is $0.300$, and the surface on which the $8.00$ -kg block rests is frictionless. $\mathrm{A}$ constant horizontal force of magnitude $F=10.0 \mathrm{~N}$ is applied to the $2.00-\mathrm{kg}$ block, setting it in motion as shown in Figure $\mathrm{P} 5.65 \mathrm{a}$. If the length $L$ that the leading edge of the smaller block travels on the larger block is $3.00 \mathrm{~m}$, (a) how long will it take before this block makes it to the right side of the $8.00-\mathrm{kg}$ block, as shown in Figure P5.65b? (Note: Both blocks are set in motion when $\mathbf{F}$ is applied.) (b) How far does the $8.00-\mathrm{kg}$ block move in the process?

A block of mass $m=2.00 \mathrm{~kg}$ rests on the left edge of a block of larger mass $M=8.00 \mathrm{~kg} .$ The coefficient of kinetic friction between the two blocks is $0.300$, and the surface on which the $8.00$ -kg block rests is frictionless. $\mathrm{A}$ constant horizontal force of magnitude $F=10.0 \mathrm{~N}$ is applied to the $2.00-\mathrm{kg}$ block, setting it in motion as shown in Figure $\mathrm{P} 5.65 \mathrm{a}$. If the length $L$ that the leading edge of the smaller block travels on the larger block is $3.00 \mathrm{~m}$,
(a) how long will it take before this block makes it to the right side of the $8.00-\mathrm{kg}$ block, as shown in Figure P5.65b? (Note: Both blocks are set in motion when $\mathbf{F}$ is applied.) (b) How far does the $8.00-\mathrm{kg}$ block move in the process?

Key Concepts

Kinematic Equations

Kinematic equations describe the relationships among displacement, velocity, acceleration, and time for objects moving with constant acceleration. They are pivotal in solving problems involving motion over a known distance or time, allowing the calculation of time durations and displacements for each body in the system under constant acceleration conditions.

Relative Motion

Relative motion analysis involves examining the movement of one object with respect to another moving object, rather than with respect to a fixed frame. When two bodies are in motion, understanding their relative velocities and accelerations is crucial for determining the conditions under which one body might reach a certain position relative to the other.

Kinetic Friction

Kinetic friction arises when two surfaces slide against each other and is characterized by a frictional force proportional to the normal force, with the coefficient of kinetic friction as the proportionality constant. It plays a critical role in dynamics problems by opposing the relative motion between contacting surfaces and is essential to calculating net forces in systems where slipping occurs.

Newton's Second Law

This fundamental law of motion relates the net force acting on an object to its acceleration, with mass as the constant of proportionality. In problems involving multiple bodies and friction, each body’s acceleration is determined by summing the forces acting on it and then applying F=ma, which helps in setting up equations for both objects in the system.

Transcript

00:01 All right, i'm going to start by drawing the diagram, f, m, m, and l.

00:10 Interesting.

00:13 So, just from the side view, this is m, this is m, this is f.

00:25 That's all supposed to be horizontal, by the way, and this is l.

00:31 Okay, fml.

00:34 Got it.

00:35 So let's read the question.

00:38 So the mass is two and the larger mass is eight.

00:44 So m is two, m is eight.

00:51 M is eight.

00:55 M is zero point three, except this is on a frictionless surface.

01:11 Okay.

01:15 F is a, is 10 newtons.

01:32 How long will it take for the small block to travel the whole way? okay, i'm not sure i've seen a problem like this before, but let's show the forces on the small block.

01:52 On the small block, we've got f going this way, we've got mg going this way, and then we've got a normal force, and then we've got a normal force.

02:05 Which opposes mg, so that would also be mg acting upward.

02:12 We also have a frictional force, which is mu times the normal force, which would be mu -m -m -g.

02:33 So, is anything else going on here? i would say no.

02:50 Okay.

02:51 So therefore, f minus mu m .g equals mass times acceleration of that one.

03:07 Okay, now let's draw a free body diagram of the larger mass.

03:13 It has a force acting downward, which is capital m times g.

03:24 It has no friction force.

03:29 It's got a force acting downward initially on it, which is over here, m times g.

03:42 It's got a normal force, and the frictional force is going to be acting in this.

03:58 Direction.

04:02 So the frictional force, which is mu m g, is going to equal the mass times the acceleration of the big block.

04:23 So b.

04:25 And this would be acceleration, a little bit confused about what that would be up there.

04:35 Nevertheless, i want to see, are there any other forces here? friction.

04:53 Okay.

04:54 So the acceleration of the big block is going to be mu little m, g over big m.

05:08 But the acceleration of the small block is going to be f minus mu m g over m over m.

05:26 Now, let's see, this is 65, so the answer should be in the back of the book.

05:35 And i'm going to head to the back of the book, chapter 5, number 65, so that i will know whether i have the right answer or not.

05:51 Chapter five in chapter seven here six okay now um let's just continue with this um l and i'm just trying to decide is this acceleration relative to the to the lower block or not so let's think about what's going on here.

06:37 We've got f going that way.

06:41 We've got the friction going backwards, mg downwards.

06:57 So it's going to accelerate.

07:01 But while it accelerates, the lower one accelerates too.

07:10 So i'm thinking that the relative acceleration is going to be f minus mu m g over m minus mu m g over m and if that's the case then the question which was what was the question how long does it take so we're looking for time um x equals x initial which is zero plus v initial t, which is also zero, plus one -half -ar, t squared.

08:24 So, and x is l.

08:29 And i don't know, i didn't write down what l is, but i'm going to erase this and write l.

08:36 So t is going to be the square root of 2l over a.

08:47 But a is f minus mu m g over m minus mu m g over m.

08:56 M minus mu m g over m.

09:03 Okay.

09:04 Do we know what l is? i don't see what l is.

09:40 So let me go back to the actual question in the book.

09:46 This is 65, and i'm going to write down the answers...