How do you find?- the spring constant.?

Question

B. The maximum compression of the spring when the mass runs in to it?

c. the mass rebounds as a result of having a compressed spring. what is i's velocity when it leaves contact with the spring?

Given:
A mass=0.70kg slides along the y-axis of a horizontal frinctionless surface with the speed Vx=2.2m/s. It runs in to a relaxed spring oriented along the x-axis. the spring had previously been observed to stretche by 3.8cm when it was oriented verticslly with the mass suspended from it.

Answer 1

Easiest is the previous observation (deflection with hanging weight). I hope that's option A. The spring constant k = f/x = g*0.7/.038 N/M.

Answer 2

B. the maximum compression of a spring = to kinetic energy is 1/2kx^2=1/2mv2. but you would need to know the mass, velocity and distance of compresssion. if you do not know that you can use hookes constant which is F=-kx

x is the distance by which the spring is elongated [usually in meters],
F is the restoring force exerted by the spring [usually in Newtons], and
k is the spring constant or force constant of the spring. The spring constant has units of force per unit length [usually in Newtons/meter

So f=mg=(.7kg)(9.8)=6.86kgm/s^2
6.86=k(.038m)
180.52=k hmm that seems really high to me.. for a spring constant

but then its plug and chug
1/2kx^2=1/2mv^2
.5(180.52)x^2=1.694kgm^2/s^2
90.26x^2=1.694
x^2=.01876
x=.137m

for such a large spring constant i am wondering if i did my eq wrong. but well i hope that clears things up for you


but in a frictionless world or perfect enviornment. energy is conserved there fore, conservation of energy and momentum laws are ineffect s yoru velocity will be the same as it leaves contact witht hte spring... 2.2m/s