# Homework Help: A man jumps on a platform supported by springs (Work and Energy)

1. Nov 13, 2009

### Chandasouk

1. The problem statement, all variables and given/known data
An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform down a maximum distance of 0.240 m below its initial position, and then it rebounds. The platform and springs have negligible mass.

What is the man's speed at the instant he depresses the platform 0.120 m?

If the man just steps gently onto the platform, what maximum distance would he push it down?

Initial (Top) PE = Final(Bottom) KE+ Uspring

mgh = .5mv2 + .5k$$\Delta$$x2

I thought it would be simple from here, but I realized I didn't have a spring constant k. I think I found it though. Since K =N/m, it'd be the same as K=mg/m ?

He compressed it it's maximum length of .240m so...

K=(9.80m/s^2)(80kg)/.240m = 3266.7N/m

So, to answer the first part, I just plug in my givens

(80kg)(9.80m/s^2)(2.50m) = .5(80kg)V^2 + .5(3266.7n/m)(.120m)^2

V = 6.96m/s ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 13, 2009

### rl.bhat

Total fall of PE of man is mg(h + x). Equate it to U(spring) and solve for k.

3. Nov 13, 2009

### Chandasouk

so, k spring would equal 2mgh/x^2

and that gives

k=285288.8889n/m

2(80kg)(9.80)(2.62m)/(.120)^2

If I plug this is though

(80kg)(9.80m/s^2)(2.62m) = .5(80kg)V^2 + .5(285288.8889n/m)(.120m)^2

I get a funky answer for V.

Last edited: Nov 13, 2009
4. Nov 13, 2009

### rl.bhat

No. What is the total energy of the man in the above case?

5. Nov 13, 2009

### Chandasouk

so, k spring would equal 2mgh/x^2

and that gives

k=285288.8889n/m

2(80kg)(9.80)(2.62m)/(.120)^2

the height is h +x which would be the 2.50m +.120m

If I plug this is though

(80kg)(9.80m/s^2)(2.62m) = .5(80kg)V^2 + .5(285288.8889n/m)(.120m)^2

I get a funky answer for V.

6. Nov 13, 2009

### rl.bhat

(80kg)(9.80m/s^2)(2.50m) = .5(80kg)V^2 + .5(3266.7n/m)(.120m)^2
The above equation of energy should be
(80kg)(9.80m/s^2)(2.50m) + .5(80kg)V^2 = .5(3266.7n/m)(.120m)^2
Now solve for v.

7. Nov 13, 2009

### Chandasouk

(80kg)(9.80m/s^2)(2.50m) + .5(80kg)V^2 = .5(3266.7n/m)(.120m)^2

1960J + 40kgV^2 = 23.52024J

40kgV^2 = -1936.47976J

V^2 = -48.411994

but u cna't take the square of a negative. Even if i neglect the sign and square it to obtain the answer 6.96m/s, my masteringphysics says it is incorrect

8. Nov 13, 2009

### rl.bhat

The k value should be
k = 2*m*g*(h + x)/x^2
Find the k value.

9. Nov 13, 2009

### Chandasouk

K=2(80)(9.80)(2.62m)/(.120)^2

K = 285282.9n/m

(80kg)(9.80m/s^2)(2.50m) + .5(80kg)V^2 = .5(285282.9 N/M)(.120m)^2

V = 1.53m/s ?

Mastering Physics told me this was wrong though?

10. Nov 13, 2009

### rl.bhat

When you want to find k value, take x = 0.240 m