What a lovely physics questions about springs...?

Question

A block of mass m is pushed against a spring that has a spring constant k. The spring is compressed by a distance d, and then the block is released. We observe that the block is launched by the spring along a horizontal frictionless surface with a final speed v.

If a second block, this one having a mass of 4m, is pushed against the spring and released, we observe that it gains a final speed of 2v. By what distance was the spring compressed in this second case?

2d
25d
16d
d
4d

(basically which factor does it go up by)... if someone could explain this to me as opposed to just giving me the answer it would be greatly appreciated.. thanks again

Answer 1

Given that m = m(0), v=v(0), and x=d, then

F = kx = ma => the spring constant k = ma/x = m(0)a/d

Using equations of motion v(f)^2=v(i)^2-2ax, we get:

a = (v(f)^2 - v(i)^2)/2x = v(0)^2/2d

Therefore, k = (m(0)/2) (v(0)/d)^2

So if now m=4m(0), and v=2(v(0), what is x in terms of d?

F = kx = ma => x = ma/k

Now using a = (v(f)^2 - v(i)^2)/2x, we get a = 2v(0)^2/x

Also, plug in the terms we solved earlier for k, then:

x = [4m(0)(2v(0)^2/x)] / [(m(0)/2) (v(0)/d)^2]

Now solve for x, and we get: x = 4d.

Answer 2

I think since it goes by a factor of 4 so the answer would be d. In otherwords it requires 4 times the mass to slow an objects v to 1/2 of an object 4 times light given the same force is applied.
Nevermind, I thought the initial speed of the 1st one was 4v, lol : )

Answer 3

Hi. I'll try. The mass of the second block was 4 times the mass of the first. It achieved a speed double the first. This is a factor of eight. I'm guessing the answer is 16d.