Newton’s Second Law is one of the toughest of the laws to understand but it is very powerful. In its mathematical form, it is so simple, it’s elegant. Mathematically it is F=ma or Force = Mass x Acceleration. An easy way to remember that is to think of your mother trying to get you out of bed in the morning. Force equals MA’s coming to get you! (I did mention how bad physics jokes are, right?)


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7 Responses to “Downhill Race”

  1. 1 Newton is a measure of force, so that’s what it would weigh. When you step on the bathroom scale, that’s the force you’re pushing down on the earth, so that’s also a weight. If you weighed 100 pounds, then you’d be pushing down with 100 pounds. There are 4.448 pounds in every 1 Newton of force.So if you weighed 100 pounds, it would be 444.8 N that you’d also weigh.

  2. Laura Swick says:

    when you said that one newton was as much force to lift a glass of milk, how heavy milk will that be?

  3. Henny van Dongen says:

    Thank you for your explanation, Aurora, this is much more clear now. But there is another thing I have difficulty understanding: In the video you say that terminal velocity starts at the end of the second second, so after 7,5″ drive. However, if I calculate the acceleration in the first second I end up with a = 2d/t^2 = 2×2,5″/1^2=5″/1=5″/s/s, then I interpret this like: at the start the velocity was 0″/s, then it accelerates with 5″/s/s/, so at the end of the first second, the velocity is 5″/s.
    If this is correct, I would say, if the toy car keeps going on a steady speed, without accelerating, it will go 5″ in the next second. This it does. I would say then, that terminal velocity start at the end of the first second, because after that the car doesn’t accelerate anymore. But since this is not what you said in the video, I need some enlightening here, please.

  4. It sounds like you’re talking about static friction versus kinetic friction. Static friction occurs between two surfaces that are not moving relative to each other, and kinetic friction occurs between surfaces that ARE moving relative to each other and rubbing against one another. In the case of the experiment with a toy car, the forces on the car when it’s at rest are (1) the static friction between the car’s wheels and the ground, (2) the static friction between the center of the car’s wheels and the axles around which they rotate, and (3) gravity. Gravity is stronger than the two static frictional forces, and so the car starts to accelerate down the ramp. As the car starts to gain speed, the static friction between the car’s wheels and the ground still exists, assuming the wheels come in solid contact with the ground and aren’t slipping along the surface. But the friction between the car’s wheels and their axles changes from static to kinetic – the centers of the wheels start to rotate around the axles and the two surfaces rub together. As the car continues to accelerate, this kinetic frictional force increases (the wheels will rub more and more strongly against the axles) until the total frictional force on the car matches the strength of gravity. As mentioned before, when these forces match strength the net force on the car is zero and it moves at terminal velocity.

  5. Henny van Dongen says:

    There is something I don’t understand. If the car stops accelerating, because the friction and gravity levels each other out, why don’t they level each other out in the first two seconds? When you release the car, the friction it has to overcome is even higher then the friction it’s under after it rides, and still gravity overcomes it. Is it because air resistance is aided to the equation?

  6. I think the confusion is coming in when you mix momentum with force. When you do the “F=ma” equation, you need to apply it to an instant in time, for example if you are standing on a bridge, you have zero acceleration, but your weight is being supported by the structure of the bridge, so the force of the bridge pushing up against your feet is in equilibrium with your weight (this is what is called statics analysis – you’ll learn about that in college).

    If you look at the moment a ball hits the wall, that’s now an impact problem in dynamics, and you have to account for the speed and the velocity of the object before and after impact using the conservation of momentum equations (again, in college!). Things get much more complicated when acceleration factors in.

    So back to your original question: a car going 30 mph has more momentum than a fly going 30 mph because the car is more massive than the fly, and it’s going to hurt more when it whacks into you.

    Momentum is calculated by mass*velocity (P=mv).
    Force = mass*acceleration (F=ma and F=d(mv)/dt).

    It’s how you use the two equations above that let you really see how it works. You “snap a picture” at two moments in time to figure out how the system changed. For the fly hitting you, take the instant before it hits you (P1 = momentum before impact = (mass)*(30mph)) and the instant after it hits you (P2 = momentum after impact = (mass)*(0mph) = 0) to find the total momentum (which is only P1 now since P2=0). You can translate the force you feel from the fly from the momentum equation by using this equation: F=dP/dt.

    Does that help?

  7. Kenneth B Newman says:

    One thing I find a little confusing about the example of a car going 30 mph and a fly going 30 mph is that if they are going at a constant velocity they are not accelerating, so acceleration = 0. So by F = m*a, F = m*0 = 0 in both cases, so the force of a car hitting an object is 0 and the force of a fly hitting an object is 0. Clearly if I’m the object getting hit, I’ll pick the fly over the car, but by this definition the Force is 0. I think my confusion might be over the definition of Force.