A 4.5 -kg sled is pulled with constant speed across a rough,...

Key Concepts

Data Analysis and Linear Regression

Data analysis techniques such as plotting experimental data and applying linear regression are crucial in physics experiments. In this context, plotting the frictional force against the normal force results in a linear relationship, where the slope of the line directly corresponds to the coefficient of kinetic friction, allowing for its determination from experimental data.

Coefficient of Kinetic Friction

The coefficient of kinetic friction is a dimensionless quantity that characterizes the frictional resistance between two moving surfaces. It is defined as the ratio of the kinetic friction force to the normal force, and is used to quantify how much frictional force resists sliding compared to the force pressing the two surfaces together.

Normal Force

The normal force is the reactive force exerted by a surface that is perpendicular to the object in contact with it. In the context of horizontal motion, the normal force is typically equal to the weight of the object plus any additional load, and it directly influences the frictional force experienced by the moving object.

Equilibrium at Constant Velocity

When an object is moving at a constant velocity, it is in a state of dynamic equilibrium, meaning that all forces along the path of motion are balanced. This allows the applied pulling force to equal the frictional force, thereby enabling straightforward calculations of friction based on measured forces.

Transcript

00:01 All right, question 61.

00:03 A 4 .5 kilogram sled is pulled with constant speed across a rough horizontal surface, with various amounts of weight added to the sled.

00:14 Data showing the required force for several different amounts of added weight are given in the table.

00:20 The question asks to plot the data and determine the coefficient of kinetic friction between the sled and the horizontal surface.

00:26 So this is our scenario here.

00:28 If the black rectangle is our sled, the blue box on top could represent the amount of that is added to it.

00:34 And like a science experiment, we can measure the force required in order to pull the sled under these new conditions.

00:41 So our job to say is to say, okay, we know this is happening.

00:48 What is the kinetic friction acting on the box or the sled in order for these measurements to occur, right? so what we do is we look at a force body diagram, sorry, a free body diagram acting on the sled.

01:06 So the sled will say with zero added mass just for simplicity you know there's a normal force acting on it has a weight m g i should state this question because we'll deal with x and y's what my axes are for the scenario so x would be horizontal the y is vertical if we're pulling the for pulling the sled i say in this direction to the right hand side then we have a posing frictional force acting to the left fk because it's a connect friction we're dealing with then we look at the net force in in our x direction, we know that the final result will be ma, because we're pulling it to the right -hand side, and that has to equal and balance our, with our coefficient of kinetic friction.

02:11 And then our forces in our y direction are simply just given as the normal minus m -g.

02:18 The weight has to be zero, because it's not accelerating in our vertical direction.

02:25 So if we look at our part a question, we say it's ma, i'm gonna write this, just as the force being pulled, because that's what we'll need for later on.

02:40 So we know the force that the sled will be pulled with is equivalent to the force to kinetic friction, which we know is always given to be the coefficient of kinetic friction times the normal.

02:50 From our y -force equation, we know the normal is just m times g, for any given mass that the sleigh might have at that time, so mu -k times m -g.

03:06 We're solving for the force of kinetic friction, sorry, the coefficient of kinetic friction.

03:14 So if you look, if we rearrange the equation, we can write that we have f over m times g.

03:26 So in the scenario here, we have this term, which for any individual point given in this figure, in the data provided to us originally, we know, right? we know force, we know mass.

03:38 However, because we're dealing with sort of like a, well, our data is being plotted scenario, we can think of it as change in force over, change in mass, which change over y over change for x is just a slope.

03:53 So if we part, the first part of this question asks us to plot our given data, which will be actually is a little comp, well, not complicated, but will not be as accurate for me to do on this type of screen.

04:07 But if you go along, you can add in either values with zero.

04:15 We'll see what every three lines is one, one, zero, one, two, two, two, three, four.

04:27 I mean, you can spread this out a bit more...

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