A man jumps on a platform supported by springs (Work and Energy)

[Chandasouk]

Homework Statement

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?


I start with an energy bar graph

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

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 ?

 

[rl.bhat]

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

 

[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.

 

[rl.bhat]

so, k spring would equal mg/2.62m =299.23n/m?

and the equation is now

(80kg)(9.80m/s^2)(2.62m) = .5(80kg)V^2 + .5(3266.7n/m)(.120m)^2

V = 7.12m/s ?

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

 

[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.

 

[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.

 

[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 = -48.411994but 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

 

[rl.bhat]

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

 

[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?

 

[rl.bhat]

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