One more question A trampoline and potential energy

In summary: So the total change in height is 3.612 meters.In summary, a 75 kg trampoline artist jumps vertically upward from a platform with a speed of 4.0 m/s. The trampoline, which behaves like a spring with a spring constant of 5.2 x 10^4 N/m, is located 3 meters below the platform. When the artist depresses the trampoline, the total change in height is H + x, where H is 3.0 meters and x is the distance the trampoline is depressed. By equating energy at the top and bottom, the artist can solve for x, which is 0.306 meters. However, this calculation does not account for the artist's
  • #1
MJC8719
41
0
A 75 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.0 m/s. the tampoline is three meters below the platform.


(b) If the trampoline behaves like a spring of spring constant 5.2 104 N/m, how far does he depress it?

when he depressed the trampoline he also moved further downward. So the total change in height is H + x where H = 3.0 meters and x is the distance the trampoline is depressed, not just 3 meters

You can equate energy at the top, when he is motionless, with energy at the bottom, when he is also motionless.

So equating energy...

m g (H + x) = (1/2) k x2

You'll have a quadratic in x. Go ahead and plug in numbers:

75 * 9.8 * ( 3 + x ) = (1/2) * 52000 x2

2205 + 735 x = 26000 x2 ugly. you can div everything by 1000

26 x2 - 0.735 x - 2.205 = 0 and use quadratic eqn

x = ( 0.735 +/- (0.7352 - 4* 26* (-2.205) )1/2 ) / 2*26

x = ( 0.735 +/- 15.16 ) / 52 = -0.277, 0.306

The negative is not valid, so the answer must be 0.306 meters
 
Physics news on Phys.org
  • #2
MJC8719 said:
when he depressed the trampoline he also moved further downward. So the total change in height is H + x where H = 3.0 meters and x is the distance the trampoline is depressed, not just 3 meters
Good.

You can equate energy at the top, when he is motionless, with energy at the bottom, when he is also motionless.
But he's not motionless when he jumps from the platform; he's moving at 4.0 m/s.
 
  • #3
Ok...I understand that I missed that but I do not know how I can account for it...Is it as simple as multiplying the mg(h+x) by 4 or is it more complicated...

Im guesssing I ignored that part since I am not to sure what to do with it lol
 
  • #4
MJC8719 said:
Ok...I understand that I missed that but I do not know how I can account for it...Is it as simple as multiplying the mg(h+x) by 4 or is it more complicated...
mg(h+x) is potential energy; why would you multiply an energy times a velocity?

Instead, realize that the initial energy is not just gravitational PE, but also KE.
 
  • #5
Whoops..

So the original equation should read:
mg(h+x) + .5mv^2 (accounting for the kinetic energy) = .5 kx^2

Then just solve for x
 
  • #6
That looks good.
 

1. What is a trampoline and how does it work?

A trampoline is a device consisting of a strong fabric stretched over a steel frame with springs attached. It is used for jumping and performing acrobatic stunts. When a person jumps on a trampoline, the springs compress and store potential energy. This energy is then released, propelling the person upwards.

2. How is potential energy related to a trampoline?

Potential energy is the energy that an object possesses due to its position or condition. In the case of a trampoline, potential energy is stored in the springs when they are compressed. When a person jumps on the trampoline, this potential energy is converted into kinetic energy, which causes the person to bounce.

3. Can the potential energy on a trampoline be increased or decreased?

Yes, the potential energy on a trampoline can be increased or decreased. When the springs are stretched further, more potential energy is stored, resulting in a higher bounce. Similarly, if the springs are less stretched, there will be less potential energy stored, resulting in a lower bounce.

4. Is potential energy the only factor affecting the bounce on a trampoline?

No, potential energy is not the only factor affecting the bounce on a trampoline. The weight and force of the person jumping, as well as the elasticity of the trampoline's fabric and springs, also play a role in determining the height and intensity of the bounce.

5. How is potential energy converted into kinetic energy on a trampoline?

Potential energy is converted into kinetic energy on a trampoline through the process of elastic potential energy. When the springs are compressed, they store potential energy. When they are released, this potential energy is converted into kinetic energy, causing the person to bounce. The elasticity of the trampoline's fabric also helps to convert the potential energy into kinetic energy by providing a surface for the person to bounce on.

Similar threads

Why doesn't this solution work? (Springs and Conservation of Energy)
    • Introductory Physics Homework Help
Replies
2
Views
796
Air Friction energy loss on a trampoline
    • Mechanical Engineering
Replies
13
Views
1K
Elastic potential energy problem
    • Introductory Physics Homework Help
Replies
14
Views
3K
Jumping on a Spring-Like Trampoline problem
    • Introductory Physics Homework Help
Replies
5
Views
10K
Catapult spring, Kinetic and Potential energy
    • Introductory Physics Homework Help
Replies
10
Views
2K
Trampoline Artist Jumps: Speed & Spring Stiffness Calculations
    • Introductory Physics Homework Help
Replies
1
Views
5K
How to find Potential Energy without distance
    • Introductory Physics Homework Help
Replies
4
Views
3K
Calculating Maximum Speed and Height of a Child on a Trampoline
    • Introductory Physics Homework Help
Replies
4
Views
2K
A really confusing question on potential energy,
    • Introductory Physics Homework Help
Replies
4
Views
1K
Motion with elastic potential energy and other forces
    • Introductory Physics Homework Help
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top