Energy needed to jump one meter

1. Jul 26, 2017

EraXon

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

So I know the formula of kinetic energy (mv^2 divided by 2). And I have a mass of 50 kg and a speed of 1 m/s will that mean I online need 25 Joules to make that jump?

Last edited by a moderator: Jul 26, 2017
2. Jul 26, 2017

person123

Yes, but your title says energy to jump one meter, which would be a different problem.

3. Jul 26, 2017

EraXon

I don't get it isn't that the energy needed ?

4. Jul 26, 2017

person123

5. Jul 26, 2017

Ibix

"Jump one metre" is very vague. What are you actually trying to do, @EraXon? Jump to a height of 1m? Or jump across a 1m gap? Or something else?

6. Jul 26, 2017

EraXon

Jump straight up 1m with a speed of 1m/s

7. Jul 26, 2017

person123

What makes you think you're going to jump up 1 meter if your initial speed is 1 meter per second? You're trying to do two different problems at the same time.

8. Jul 26, 2017

EraXon

Don't know.I'm just trying to repeat stuff for high school .

9. Jul 26, 2017

person123

So there are two different problems. One was to find the energy required jump with an initial velocity of one meter per second. You found the solution for that.

The other problem is to find the energy to jump straight up to a max height of one meter (assuming the mass is 50 kg). For that you need the acceleration due to gravity. That allows you to find the gravitational potential energy.

10. Jul 26, 2017

EraXon

Yeah I know about it potential energy (mgh) but that is way bigger then the kinetic one. That's what I don't get.

11. Jul 26, 2017

person123

If you do two different problems, there's a good chance the solution to those problems are going to be different. Jumping straight up one meter requires a lot more energy than jumping with an initial velocity of one meter per second.

12. Jul 26, 2017

Ibix

You should really post this kind of question in the homework forums. I'll ask for it to be moved.

Do you know the formula for gravitational potential energy? Do you know what conservation of energy is?

Ask yourself a question: if you want to throw a ball so that it goes up 1m, what do you do? What do you do differently if you want it to go up 2m? 10m?

13. Jul 26, 2017

Staff: Mentor

Could you please clarify this? You are about to go into high school, and are trying to start doing high school physics problems? Or you are out of high school, and going back through old problems to try to learn high school physics?

14. Jul 26, 2017

EraXon

Going to high school from September it's my first high school year and I never truly understood this things. Saw this site and decided to give it a try.

15. Jul 26, 2017

EraXon

The formula is mass*gravitation acceleration * height. Conservation of energy means that energy tends to conserve itself by transforming in different types:letting a ball go that is higher than the ground it's gonna transform its potential energy into kinetic energy. I think

16. Jul 26, 2017

Staff: Mentor

I'll go ahead and move this thread to the schoolwork forums now. That's the best place to ask schoolwork-type questions, and when you start a new thread there, you are provided with a Template that asks for the Problem Statement, the Relevant Equations, and your Attempt at a Solution. In your future threads, use that Template to help you organize your question, and to show others what you know about the equations and how you start the problem.

For now, it may help you to browse a bit through the Hyperphysics website, to see more about the equations you can use for this type of problem:

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

(and click on the Mechanics bubble).

It sounds like you are a little familiar with the equations for gravitational potential energy and kinetic energy:

$$PE = m g h$$
$$KE = \frac{1}{2}m v^2$$

How about F = m a ? Are you familiar with how that enters into your question?

17. Jul 26, 2017

EraXon

Yeah through the L=F×d

18. Jul 26, 2017

haruspex

I do not see how that is an answer to berkeman's question. In fact, I'm not sure what your L is there.
F I take to be force, and d distance. Force times distance turns up in two standard equations but meaning quite different things.
If the force and the distance are in the same direction then their product can be the work done by the force as it advances that distance.
If the force and distance are at right angles then their product is the torque exerted by the force about an axis that far from its line of action. Your use of L suggests to me you are thinking of torque, not work.

F.d as work done does relate to PE = mgh. If a mass m is raised height h against the gravitational force mg then the work done is mgh.
If you want to know the take-off speed for a jump that reaches height h then you would write initial KE = gain in PE, i.e. ½mv2=mgh.