# Spring and Rotational Question

1. Aug 5, 2010

### Mercfh

1. The problem statement, all variables and given/known data
A 0.2 kg mass is held against a spring with spring constant k=1000N/m. It is launched from the spring
and travels an additional 1 m across a surface with coefficient of kinetic friction µk=0.2. It then collides
and sticks to a 0.3 kg mass that is suspended from a 0.5 m long thread of negligible mass. How far back
must the spring be pulled (∆x) in order that the combined masses swing fully around the support pivot?

2. Relevant equations
Velocity Min to get around a loop= sqrt(5gR)
Spring Distance Launched= d = Kx^2/(2mugm)
Velocity of a Spring? = V^2=(k/m)*d^2
Distance = x = V^2/2a
Acceleration = m*g

3. The attempt at a solution
Well I wasn't sure if the Velocity equation was the "right" equation to find this. ideally I would "think" you would find the minimum speed that the spring would launch.....but this doesn't put in mu for the velocity, so im not sure how you calculate that. But once you find the velocity after 1m I figure you'd just check and see if that meets the minimum velocity to spin the object around the pivot.
I could be COMPLETELY wrong however

2. Aug 5, 2010

### arkofnoah

i don't think you should use the escape velocity equation (??) to do this. hint: find the gravitational potential energy gained :D

Last edited: Aug 5, 2010