Calculating Spring Constant for Large Springs and Trampolines

[Sally99]

Homework Statement

You are making a large spring. You want a 150kg person to be able to stand on the spring an be above or just touch the ground which is 15 cm below.

a) What is the minimum spring constant the large spring should have?

b) From the above value, if you have a mass of 70kg and land on the trampoline with a velocity of 3m/s how far do you drop? Will you touch the floor?

Homework Equations

F= -kx

The Attempt at a Solution

a) The force of the man is equal to Fg. So 150kg x 9.8. Using this force we could plug it into the equation F= -kx (x is 0.15m) to solve for k? Does that seem right guys?

b) The force of a 70kg person would be Fg which would be 9.8 x 70. But then you also need to factor in that the velocity would add to the force and then use Hooke's law F=-kx. I'm not sure how the velocity gets factored into this equation?

Anyone think I'm on the right track/ know how to do this?

 

[PhanthomJay]

Homework Statement

You are making a large spring. You want a 150kg person to be able to stand on the spring an be above or just touch the ground which is 15 cm below.

a) What is the minimum spring constant the large spring should have?

b) From the above value, if you have a mass of 70kg and land on the trampoline with a velocity of 3m/s how far do you drop? Will you touch the floor?

Homework Equations

F= -kx

The Attempt at a Solution

a) The force of the man is equal to Fg. So 150kg x 9.8. Using this force we could plug it into the equation F= -kx (x is 0.15m) to solve for k? Does that seem right guys?

I don't think so. It would be correct if the man was allowed to be slowly lowered onto the spring say by someone not on the spring holding onto him as he is lowered to the equilibrium position. However, the problem is not worded so good, but I interpret it as the man standing on the spring unassisted, in which case you must apply the conservation of energy principle to solve for k.

b) The force of a 70kg person would be Fg which would be 9.8 x 70. But then you also need to factor in that the velocity would add to the force and then use Hooke's law F=-kx. I'm not sure how the velocity gets factored into this equation?

Anyone think I'm on the right track/ know how to do this?

again, use conservation of energy principle...are you familiar with it?

 

[Sally99]

Yeah and thanks for the reply! I actually used that equation for part b).

But if i did use the conservation of energy principle wouldn't I need to be given velocity?

 

[denverdoc]

Velocity is given: 3m/s. That kinetic energy and potential energy (relative to the final stop point) is converted to spring compression whose energy is given by 1/2kY^2.

so 1/2 Mv^2+mg(y)=1/2(Ky^2).

 

[Sally99]

The velocity in this question is only meant for part b) though.

 

[denverdoc]

The velocity in this question is only meant for part b) though.

Right. The spring constant is computed from the first question.

 

[ideasrule]

a) The force of the man is equal to Fg. So 150kg x 9.8. Using this force we could plug it into the equation F= -kx (x is 0.15m) to solve for k? Does that seem right guys?

Yup.

b) The force of a 70kg person would be Fg which would be 9.8 x 70. But then you also need to factor in that the velocity would add to the force and then use Hooke's law F=-kx. I'm not sure how the velocity gets factored into this equation?

Unless you know how to solve differential equations, you pretty much have to use the conservation of energy to solve this problem, as the other helpers mentioned.

 

[Sally99]

Ok that's what I did, thanks for your help guys!

 

[PhanthomJay]

The velocity in this question is only meant for part b) though.

in part a , the initial velocity is 0.