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February 5, 2015 · by physics · Tension, Tension physics

Question:

An 1130 Kg car is held by a cable on a smooth frictionless ramp.  Cable makes angle of 31 degrees above surface of ramp.  The ramp rises 25 degrees higher than the horizontal.  Find the tension in the cable.  

Transcripts

Hello, everyone. This is Tom from everystepphysics.com and everystepcalculus.com. Today, we’re going to do everystepphysics.com stuff. A 1100 kilogram car is held in, oh they have plane here but it’s place. A 1100 kilogram car is held in place by a light cable on a very smooth frictionless ramp as shown in the figure. The cable makes an angle of 31 degrees above the surface of the ramp and the ramp itself rises 25 degrees above the horizontal. Note that the magnitude of gravitational acceleration of the earth is 9.8 meters per second squared. Draw a free body diagram for the car and find the tension in the cable. So how do we do this problem? Well, you go to physics to get to my menu. Then you choose tension because that’s what it’s asked for in the problem. So we go up tension, press Enter. And we have all these choices. We’re given to angles and one mass. In this case, we would choose number three. We put in there the mass of the car.Alpha 1130 kilograms.
Again, the anglesAlpha 31 degrees plus 25 degrees.We got 56.Notice if the ramp angle was given, you’d be pulling just along the frictionless ground. And so that there’s only one angle given, that would be mg times sin of theta. Which is a standard tension formula. Now if you add the car going with it going up a ramp with a angle, then and the rope is pulling on car at an angle, you have to add those angles together which is 56. Now you do the calculations. Equals 9190.1 newtons. Okay. Sometimes you have to kind of..one great thing about my programs is it gives you the formula for tension. A lot of times, a really simple simple formula, you can’t even recall them because you really don’t don’t have any way of placing them together and using them. My programs give you at least the formula for tension.

February 3, 2015 · by physics · Bullets, Velocity in Physics

Question:

A pendulum consists of a 2 kilogram block hanging on a 1.5 meter
length of string. A ten gram bullet moving with a horizontal velocity of nine hundred meters per second strikes passes through and emerges from the block which was initially at rest with the horizontal velocity of three hundred meters per second. To what maximum height above its initial position will the block hang?

Transcripts

Hello, everyone. I’m Tom for everystepphysics.com and everystepcalculus.com. I’m going to do physics problem here. I’m going to read it right now. A pendulum consists of a 2 kilogram block hanging on a 1.5 meter length of string. A ten gram bullet moving with a horizontal velocity of nine hundred meters per second strikes passes through and emerges from the block which was initially at rest with the horizontal velocity of three hundred meters per second. To what maximum height above its initial position will the block hang? So let’s do it and show you how my programs work on this. You enter physics here to get to my main menu. We’re going to scroll down to bullet and pendulum. Which is right. And we’re going to enter our variables. You have to press enter before you enter anything in these entry lines here. So it asks to enter mass so we’re going put Alpha 3 and grams. I show you how to convert that to meters. And second mass is Alpha 2 kilograms. Initial Velocity is Alpha 1000 meters per second. And this is at Alpha 0 because the block is a rest hanging. And then we’re going into the final velocity is the bullet passes through Alpha 300 meters per second. And given in the positive direction east. Enter length of string is Alpha 1.5. I don’t know what they gave that to you in a problem for because it’s not entered into the calculations so must be to throw you off or something which professors love to do. This is at meters. So, I show you what you’ve entered, you can change it if you want. I say it’s okay. And it’s asking for the height so we’re going to choose number 3 here. And you have to find the velocity first. Which is .75 meters per second squared. I mean per second meters per second. And then we use that velocity to find the height which is .02867 meters. Centimeters would be .029 close to the 27 centimeter answer And that’s what you’d circle and get a 100 percent in this problem. You can choose 2 also and get to the and find the actual velocity of the block. Which is .75 meters per second. Pretty neat, huh? everystepphysics.com. Thank you.

November 11, 2014 · by physics · Modified Atwood Machine

Raw Transcript

Hello, everyone. This is Tom from everystepphysics.com and everystepcalculus.com. I’m going to do a Modified Atwood Machine problem. And that’s where the mass instead of hanging in both size and the pulley with a rope, its one mass is on a table. Interesting note, notice that the angle is 90 degrees in the corner. So as important as we in another video use friction and and angle. And also friction on the modified atwood machine but let me show you how we do this problem with my programs. Physics to get into my menu. Press enter. We’re going to scroll down to Atwood Machine. Find the acceleration. These arrows here show that there’s more on the menu in case it comes in the middle like this. And we’re gonna choose not atwood machine but modified atwood machine there. Number 3. You have to press Alpha before you put anything into these entry lines here. and the m1, mass 1 of the problem is Alpha five kilograms. If it wasn’t, I would convert it for you and show you that so you could mark it on your paper. And the second mass, m2 is alpha 7 kilograms, again. Choose number two. I show you what you entered in case you made a mistake. I say it’s okay. Here’s a picture just like in case you needed to the draw the picture yourself, you could use that. Here’s the formula. Really simple, isn’t it? Yeah but try to remember that formula from the way professors describe it on the blackboard and how they use the diagrams etc. And so the answer is 5.722 and they have 5.8 meters per second squared on me problem. I always liked put 5 decimal points, not 5 decimal points but 5 integers so that you can, some tests require that. And you can use 3 or 2, whatever you want. But notice they got the free body diagrams going and try to decipher that and put that into a test or homework. It takes me a long time to to do it myself when when I finally get the formula is here. And especially with angles and with friction force. And also I always do tension. That’s a big deal because it uses the acceleration that you’ve just found to find the tension. So here’s the tension 28.613 Newton’s. Pretty neat, huh? everystepphysics.com. Go to my site, buy my programs pass your physics class and subscribe to me so you can see other videos that I might make. Thank you.

· by physics · Modified Atwood Machine

September 30, 2014 · by physics · Modified Atwood Machine, Physics Videos

Raw Transcript
Hello, Everyone. Tom from everystepphysics.com and everystepcalculus.com. We’re going to do A Modified Atwood Machine. This time with friction on the table there. And you’ll notice the ninety degree angle. And that angle goes from when the two masses are hanging and he said the pulley that would be zero angle and then up at a table will be ninety degrees so on all the angles in between, I’m always bothered with physics formula’s where they don’t have a complete formula. And it’s not clear. So anyway, the cosine is in the formula and cosine of ninety degrees is zero. So we see how that works. So that’s the only thing to remember that it has an angle and UK also. So we’re gonna get going here. Physics, go to my menu. I’m going to scroll down to Atwood Machine, here. And I’m gonna scroll up to Modified Atwood with theta and UK. I’ll show you a picture in case you didn’t know how to draw it. Now, of course this would be ninety degrees table. I have that in the modified section of this program, also. I’ll show you that. We’re going to put in a mass here. M1 is equal to 5. You have to push Alpha before you enter anything in these entry lines here. So it’s Alpha 5 kilograms and then M2 is Alpha 11 kilograms. Theta then would be ninety degrees and UK would be point 8 as it’s given. I always show you which have entered, you can change it if you want. I say it’s okay. Here’s the formula. Notice UK’s in the second part of this equation. All divided by the total masses. And then we have cosine here. They enter the variables. And here’s the answer: 6.744 meters per squared. You notice how the second term of this formula became 0 because cosine of ninety is 0. Pretty neat, huh? everystepphysics.com and everystepcalculus.com. Go to my site, buy my program, pass your classes, and subscribe if you wanna see more movies. I also give tension here. See the tension because tension uses the acceleration that we found previous. Have a good day!

Raw Transcript

Hello, Everyone. This is Tom from everystepphysics.com. And I’m going to do an Atwood Machine problem. Show you how simple it is with my programs in the Titanium. And remember these are test questions. In other words, you’re going to get a test question like this and given certain masses and what are you going to put on your test? Are you going to put in the first diagram here? They want you to make a free body diagram that you’re gonna have to do when your own brain or own head. When it comes time to find the acceleration for this system,my programs as where it’s at. So let me do that for you right now. Physics to get to my menu. Physics menu. And we’re going to scroll down to at Atwood machine. And you notice these arrows go up and these go down so that means there’s more things in the menu. We’re gonna go up here to Atwood Machine. And you have to press Alpha before you enter anything in here. Notice that they give you the mass of one is 15 kilograms so we’re gonna press Alpha 15. Choose kilograms. Second mass is 28 kilograms. Alpha 28 kilograms. I always show you what you’ve entered so you can change if if you want. I say it’s okay. Here’s the little blurb on it, what it is. 2 masses hanging by a pulley. I show you a picture of it. And here’s the formula. Acceleration equals G times m2 minus G times m1. And summing the massesI do the calculations for you. Add it up, all step-by-step and here’s the answer. Acceleration equals 2.96. He has 2.977 meters per second squared. In the next problem, he solves for tension and I do that also in my Atwood machine. Notice we already have the answer here. Here’s the formulas and the step-by-step solutions 191.64 Newtons. He’s got 190. So you get an A on this problem on tests. You’re done within one minute and you’re onto the next problem. Pretty neat, huh? everystepphysics.com Go to my site, buy my programs. Pass your physics or calculus class. And subscribe to me see more movies like this. Thank you.

September 28, 2014 · by physics · Physics Test Questions

Question:

A block is pushed up a frictionless 30 Degree incline by an applied force as shown. If F=25 N and M=3.0kg, what is the magnitude of the resulting acceleration of the block?

Raw Transcript
Hello, Tom from everystepphysics.com and everystepcalculus.com. A mass on a ramp problem, test problem. Let’s do it. You have to put physics in the entry line of the calculator to get to my menu. And because the problem ask for acceleration, I always recommend you circle what’s asked for and then on the side, put down what’s given. In this case force Newtons and theta and then mass. So right now we’re gonna go to, it asks for acceleration. Press number two here. And then we’re going to scroll down to what’s given because there’s many of those in here. It’s not mass, and theta, and friction force. It’s mass, theta, and newton. There we go,letter E. You have to press Alpha before you enter anything in these entry lines. Alpha 3 kilograms. And then the force is alpha 25. And the angle given is 30. Alpha 30. I always show you what you’ve entered, you can change it if you want. I say it’s okay. Here’s the formula. Write this down on your paper right away. I like to use A equals acceleration because a lot of times and most the time in Yahoo or wherever you search, you don’t know what they’re asking for you don’t know what this letter means. Professors and people do not put that down. I like to put that down first in my problem so we all know what we’re talking about, exactly not almost. And here’s the to answer here: 25 times cosine at thirty minus 3, which is the mass times gravity, etc. divided by three, all the quantity, divided by three. Here’s the step-by-step 2.3119. You got an A on this problem. Go to my site, buy my programs everystepphysics.com. Pass your classes and subscribe if you wanna see more movies. Thank you.

September 26, 2014 · by physics · Displacement, Physics Test Questions, Physics Videos

A cart is given an initial velocity of 5 meters per second and experiences a constant acceleration of 2 meters per second squared. What is the magnitude of the carts displacement during the first six seconds of it’s motion?

Raw Transcript
Hello, Everyone. Tom from everystephysics.com. Here’s a test problem. I’m gonna read for you. A cart is given an initial velocity of 5 meters per second and experiences a constant acceleration of 2 meters per second squared. What is the magnitude of the carts displacement during the first six seconds of it’s motion? I always want you to in these problems to circle what’s asked for. In this case displacement and I want you to write on the side of the program what’s given. Which is in this case initial velocity, acceleration, and time. So, let’s the do it. You have to put physics in here with the open/close parentheses to get my menu. We’re gonna scroll down to displacement because that’s what it’s asked for. And we’re gonna choose what’s given which is velocity, acceleration, and time. There’s a formula for it. You have to press Alpha before you enter anything in these entry lines. So the the initial velocity is Alpha five and meters per second. I do the conversions. Is acceleration given? Yes. Acceleration alpha 2. And the time is alpha 6. I always show you what you’ve entered. You can change it if you want. I say it’s okay. Here’s the answer: letter C

· by physics · Acceleration, Physics Help

A European sports car dealer claims that this product will accelerate at a constant rate from rest to a speed of 100 kilometers per hour in eight seconds. What is the speed after the first five seconds of acceleration?

Raw Transcript
Hello, Everyone. Tom from everystepphysics.com and everystepcalculus.com. I going to do another physics problem. I’ll read it. A European sports car dealer claims that this product will accelerate at a constant rate from rest to a speed of 100 kilometers per hour in eight seconds. What is the speed after the first five seconds of acceleration? So, let’s do it. Physics to get into my menu. Open closed parentheses. That tells a calculator you want to do a program. And we’re going to go because I want you to circle things, what’s asked for in the problem and I want you to write on the right, I want to write was given so you keep clear mind of everything. In this case was speed. Speed velocity is the same. Just for your information, velocity is a vector can be negative or go in any direction. Speed is a scaler and it’s only positive, you can never have it negative. But otherwise they’re essentially the same. So we’re going to, I’m going scroll up to go to the bottom on the menu here and I’m going to scroll up to speed. Right there. And then we choose what’s given. Well and what’s given is final velocity in time. So I’m gonna go up in the menu here and see if that’s in here. And here’s velocity and time, right here. Choose that. They’ve given you initial velocity or final velocity. So you press Alpha, press Alpha first before entering anything into the entry lines. Alpha 100. And notice they give you kilometers, change the meters per second. So I do that for you. You write this on your page do the conversion. And is acceleration given? No. No acceleration is given. Okay and then Alpha. Alpha and what time was given. Eight seconds. So then I do the acceleration for you here it is right here. And then is another time given. I say yes. So I’m going to go Alpha five. And here’s the answer for the test problem. 17.4 meters per second. Pretty neat, huh? everystepphysics.com Go to my site, buy my programs and maybe subscribe for more videos. Thank you

September 24, 2014 · by physics · Displacement, Physics Videos

A rock is thrown straight down with an Initial Velocity a 14.5 meters per second, from a cliff. What is the rocks displacement after two seconds? Acceleration due to gravity is 9.8 meters per second squared.

Raw Transcript

Hello, Everyone. Tom from everystepphysics.com I’m going to read a test problem to you. A rock is thrown straight down with an Initial Velocity a 14.5 meters per second, from a cliff. What is the rocks displacement after two seconds? Acceleration due to gravity is 9.8 meters per second squared. So let’s get started. Physics to get my menu. Scroll down to Displacement because that’s what you’ve circled. And that’s what is asked for the problem. And we scroll to what’s given and in this case velocity. Initial velocity, time, and acceleration. There’s a formula. Write that on your paper. And you have to press Alpha before enter anything in here. So we have to press Alpha. Initial Velocity is 14.5 meters per second. Is acceleration given? It is but in case it wasn’t, I’m going to show you this, you can put 9.81 in there but if you didn’t, it’s assumed that you’re supposed to know that. So I’m going to choose number 2. It goes right away to time. Alpha is two seconds. And I always show you what you’ve entered so you can change it if you want. I say it’s okay. Here’s the answer: 48.6 And they have for their answer 49 when they here for their answer 49. You’d circle that one and get an A on this problem. Pretty neat, huh? everystepphysics.com Go to my site, buy my programs, and subscribe. Thank you

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