Question

A 3kg object is released from rest at a height of 5m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant k=300 N/m. The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. What is the distance x?

Answer 1

Any ideas?

Answer 2

Well, Solution for the spring force constant is:
\[k=F _{x}\div(x-x _{0})\]

Where \[k\] is the Spring Force Constant, \[F_{x}\] is the Force, \[x\] is distance from Equilibrium, and \[x_{0}\] is the Spring Equilibrium Position.

So, if we wanted to get the Distance, it would be:
\[x=x _{0}-(F _{x}) \div k\]

The problem is, you have a given mass, and not the Force. If we wanted a Force from the mass, we don't have the acceleration value because F = mass x acceleration.

Let's see if we can get a workaround. There's also an issue of the Curved Ramp. I don't know if it will affect the distance and time.

(It's been a long time since I had Physics.)

Answer 3

I think the potential energy would be the force, wouldnt it?

Answer 4

I know potential energy has something to do with it but i dont know what :/

Answer 5

The object first has gravitational potential energy, call that GPE.

Then at the bottom of the ramp just before it touches the spring all of that GPE is converted into kinetic energy KE.

Then all of that KE is converted in spring potential energy, call that SPE.

In short, at the end of the process

SPE = GPE.

Now, write down expressions for both the GPE and SPE, then set then equal and solve.