You are a Starfleet captain going boldly where no one has gone before. You land on a distant pla

step 1 Hello, my name is David. In this video, we'll cover Newton's second loud motion. So for this pro

You are a Starfleet captain going boldly where no one has...

Key Concepts

Newton's Second Law

Newton's Second Law states that the net force acting on an object equals the mass of the object times its acceleration. This fundamental principle allows us to relate the forces acting on a system to its acceleration, forming the basis for setting up equations that can solve for unknown quantities such as the mass or gravitational acceleration.

Free-Body Diagram

A free-body diagram is a graphical illustration used to visualize all the forces acting on a single object. By representing forces such as an applied force and gravitational force, a free-body diagram aids in setting up the correct equations based on Newton's Second Law, ensuring that all contributing factors are considered during analysis.

Graphical Analysis of Linear Data

Graphical analysis involves plotting experimental data to reveal underlying relationships between variables. In this context, plotting the applied force versus the acceleration results in a straight line, where the slope can be interpreted as the mass of the object and the intercept relates to other forces such as the weight. This method of analysis is a powerful tool for deducing physical parameters from measured data.

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the associated forces. It involves relating quantities like displacement, time, and acceleration through equations of motion. Determining acceleration from the time it takes to move a known displacement is a common kinematic calculation used in experimental setups.

Experimental Error

Experimental error encompasses the deviations and uncertainties inherent in measured data. Variations in measured values can arise from limitations in instruments, environmental conditions, or random fluctuations. Recognizing and accounting for experimental error is crucial for accurately interpreting data and applying methods like regression analysis to extract reliable physical parameters.

Transcript

00:01 Hello, my name is david.

00:02 In this video, we'll cover newton's second loud motion.

00:05 So for this problem, we want to plug the force versus the acceleration, but the data only covers the force and the time.

00:13 So in order to find the acceleration, we must find the velocity first.

00:17 So let's recall that the velocity is equal to the changing distance over the change in time.

00:24 And the distance given or the upward motion is 8 .00 meters.

00:29 So to find the velocity for the first pair of values, we're going to divide 8 .00 meters divided by 3 .3 seconds, which gives us 2 .42 meters per second.

00:47 And now to find the acceleration, we must know that the acceleration is equal to the changing velocity over the changing time.

00:54 So we're going to divide 2 .42 divided by 3 .3 seconds, which gives us 0 .73, 462 meters per second square.

01:07 And we're going to do the same for all the data values given.

01:11 So now for the next one, it's just going to be 8 divided by 2 .2 .2, which gives you 3 .6 meters per second.

01:24 I mean, 3 .63 meters per second.

01:28 Okay.

01:29 And if we find the acceleration, we get that the acceleration is 1 .65289.

01:38 And if we do that for the rest, we should get 4 .7 0 .59 meters per second for the velocity.

01:49 And the acceleration will be 2 .76 818 meters per second square.

02:00 For the next one it will be 5 .3 meters per second and the acceleration is going to be 3 .5 meters per second square.

02:16 And for the last two values we get 6 .1538 meters per second and 4 .73 369 meters per second squared.

02:39 And for the last one we get for the velocity 6 .6 meters per second and 5 .5 meters per second square for the acceleration.

02:51 Okay, so now we could plot the force versus the acceleration so for recording the system since our highest value for the acceleration goes up to 5 .5 we could have six operations so this is going to be the acceleration in meters per second squared and the y -axis is going to be the force in newton's so when the acceleration is 0 .73 462 or forces 250, so this is going to be 250, 300, 350, 350, 450, 400, and 500, and the acceleration will go from 0 to 6.

04:03 So this is going to be 1, 2, 3, 4, 4, 4, 5, and 6, since our highest value for the acceleration was 5 .5.

04:11 So when the acceleration is 0 .73 462 or forces 250 newtons so 0 .73 will be somewhere around here so i'm actually going to mark it with green so 0 .73 will be somewhere around here and we'll go up to 250.

04:31 Okay, so now when the acceleration is 1 .65 or force is 300 newton so 1 .65 will be somewhere around here and we'll go up to 300 newtons now when the acceleration is 2 .76, so 2 .76 is somewhere around here.

04:50 Our force is actually 350 newtons.

04:55 Now when the acceleration is 3 .5, so 3 .5 is right here.

04:59 Our force is 400 neutons.

05:05 And when the acceleration is 4 .7, so 4 .7 should be somewhere around here.

05:10 Or force is going to be 450, so right here.

05:18 And with the acceleration is 5 .5, so 5 .5 is right here.

05:21 Or force is going to be 500 newtons.

05:26 Okay, so let's draw a best fit line, touching as many points as we can.

05:36 So the best fit, touching as many points as we can, it'll be like right there.

05:46 So now let's choose two random points on this best fit line and find the slope of the line using those two points.

05:56 So i'll say let's go above two meters per second square.

06:00 So we go above above two meters per second square, which is this point right here.

06:12 And we could also go above 5 meters per second square.

06:15 So it would go above 5 .1 meters per second square.

06:18 It will be somewhere around here.

06:21 So we eyeball those values.

06:25 We could say that this is going to be 2 .320, about 320.

06:36 And the coordinate for the second point, if we eyeball it, we could say that it's about 475.

06:53 Okay, so let's find the slope using those two points.

06:58 So let's recall that the slope and this is actually for part b the slope is actually for part b and the grab is part a okay, so let's find the slope.

07:25 So let's recall that the slope is equal to y2 minus y1 over x2 minus x1.

07:37 So we plug in our numbers.

07:39 We'll get 475 on top minus 320 divided by 5 minus 2 which is 3 so we get 155 divided by 3 and this is in newton's on top and meters per second square on the bottom so the units will be kilograms so it happens to be the mass so the slope or the mass it's equal to 51 .6 kilograms again because the newton is equal to kilogram times meter per second square and we divide by by meter per second square the meter per second square will cancel out so the units will be just kilograms so now we run this to two significant figures just like the information given on the problem the mass is actually 52 kilograms okay so now let's go ahead and find the force of gravity at the surface of the planet so we're gonna apply newton second law motion for this so we know that the net force is equal to the mass and the acceleration.

09:05 So the force is acting on the block is actually the force applied to the free end of the string minus the weight, which is equal to the mass size of acceleration.

09:20 And we know that the weight is equal to mass size gravity.

09:23 So the force minus the mass size gravity is equal to the mass tens of acceleration.

09:32 So we saw for g algebraically, we get that g is equal to f over m minus a.

09:46 So this is the formula we're going to use to find the gravity at the surface of the planet.

09:52 So using the first coordinate point, so using 0 .73, 462, meters per second square, and 250 newtons, and the mass before we rounded to two significant figures, so 51 .6 kilograms, we will plug that into our formula for gravity...

Please give Ace some feedback