Question

A massless spring of length 0.340 m is in its relaxed position. It is compressed to 78.0 percent of its relaxed length, and a mass M = 0.240 kg is placed on top and released from rest. The mass then travels vertically, taking 1.50 s to reach the top of its trajectory. Calculate the spring constant.

Answer 1

you'll need to use 2 kinematic equations, hooke's law and summing the forces in the y-direction.
(since you aren't online, i can make sure you do each step correctly, but I can just give you a vague idea of what you need to do; which is why my response is mostly talking concept and methods, not number)

we have F = kx, we know x, we need to find k. so that means we need to use F to find k.

recall
F=m*a

we know m, we need 'a'. 'a' will be a the result of the two kinematic equations:
first use
v = v + at to find initial velocity of the mass as it leaves the spring. by using the points of the peak and take off
then once you have v, use
v^2 = v^2 +2as at the point where the spring is compressed to when the object leaves the spring at take-off. were the left v is equal to the v from above and the right v is zero, and s is the compression of the spring. and solve for 'a'

now that you have 'a' , sum the forces in the y-direction:
F=F
ma=kx
we know m, we just found a, we know x, so solve for k.

hint: k will be between 350 and 450 ^_^

Answer 2

so @DemolisionWolf first I used \[V _{0} = V + at \] which I solved to be 14.7m/s then used \[\Delta Y = V _{0}t + (0.5)(a)(t)^2\] and solved that to be 11.025m which is the height

Answer 3

14.7 is correct, but we then need to use
v^2 = v^2 +2as
where it like this:
14.7^2 = 0^2 + 2*a*(.78*.34)
see it? we need the acceleration the spring puts into the object, this 'a' is then used in:
ma=kx,
then solve for k
^_^

Answer 4

oh alright I see thank you very much

Answer 5

ur welcome, I hope it makes sense ^_^