Practice Problem: Kinetic and Potential Energy of a Ball on a Ramp


Since you’re in the habit of building little
bits of physics apparatus at home, you decide to build another ramp, but this one is for
a ball to roll down. At the bottom, the track curves back up, launching
the ball into the air. If the ball you use has a mass of 0.15 kilograms,
and the ramp ranges from three meters off the ground at the top to half a meter off
the ground at the lowest point in the path of the ball, what is the velocity of the ball
at the lowest point of the ramp, and when it hits the ground after the jump? To answer this question, we need to know about
kinetic and potential energies, so if that sounds unfamiliar, check out my tutorial on
this subject in the classical physics series. Otherwise, see if you can figure it out. As we said, the name of the game here is kinetic
and potential energy. Kinetic energy is the energy of motion, and
potential energy is the energy of location. At the top of the ramp, before you release
the ball, it’s not moving, so it has zero kinetic energy, but it does have some potential
energy. To find out exactly how much, we use the equation. We can plug in the mass, acceleration due
to gravity, and the height of the ramp, and we get 4.4 kilograms meters squared per second
squared, or 4.4 joules. Now what we also have to understand, is that
as the ball moves down the ramp, the total mechanical energy is always conserved, meaning
the sum of the kinetic and potential energies, in this case 4.4 joules, is constant. So as the ball gets closer to the ground,
potential energy is being converted into kinetic energy. That also means that the instant the ball
hits the ground, where it has zero potential energy, it must have converted all of its
potential energy into kinetic energy, so the kinetic energy of the ball at that moment
must also be 4.4 joules. Now we can use that to find the velocity of
the ball at the moment of impact. 4.4 kilograms meters squared per seconds squared
is equal to one half the mass times velocity squared, so we simplify a bit, take the square
root, and 7.66 meters per second is what we get. Now for the lowest point of the ramp. Let’s get the potential energy first, which
will be the same mass and acceleration as before, but the height is now half a meter. That gives us a potential energy of 0.74 joules. Kinetic and potential energies must add up
to 4.4 joules, so the kinetic energy at this moment must be 3.7 joules. With that, we just plug into the same equation
as before, rearrange and take the square root, and we get 7.02 meters per second at that
point. Not too bad for a little garage-style physics,
if you ask me.

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21 thoughts on “Practice Problem: Kinetic and Potential Energy of a Ball on a Ramp”

  1. goincrazy11 says:

    Thanks professor Dave!

  2. harsh singhal says:

    it shouldn't be 7.02. it is exactly 7. check it again

  3. pattissiere says:

    What about centripetal velocity? Would it be the same as the calculated one through conservation of energy?

  4. Ash D says:

    Thanks so much, Professor Dave!

  5. Samyak T says:

    Thank you very much sir.

  6. Zaresh Anil says:

    Great professor

  7. Dennis Geurts says:

    I still dont get it. I need to see all the math equations.

  8. Riccardo Supino says:

    ty daddy

  9. Farzana Tahi says:

    Thank you ๐Ÿ˜Š

  10. Cave Linguam says:

    That was fun. Thanks. I love energy

  11. Dear math says:


  12. Mary Nd says:

    what about the rotational kinetic energy? do we consider it here?

  13. EESHAN PANDEY says:

    Most satisfying video. Thank you so much.

  14. Lucas Podiak says:

    This is so hot

  15. Maze01 says:

    God i love this shit

  16. Pokemon Trainer says:

    You should've wrote the solutions. Of some of them..
    So I can easily get it๐Ÿ˜ข

  17. Charlie Little says:

    Thank you Dave

  18. Janampelly Vamshi krishna says:

    Super sir

  19. Theoretical Hub says:

    Thanks dad

  20. Abdul Aziz says:


  21. cherry blossom girl says:

    thank you๏ผ

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