Jump to content

Recommended Posts

Posted

I am stuck on part b of this question.

 

2. A 0.20 kg mass is hung from a vertical spring of force constant 55 N/m. When the spring is released from its unstretched equilibrium position, the mass is allowed to fall. Use the law of conservation of energy to determine

a)The speed of the mass after it falls 1.5 cm

b)The distance the mass will fall before reversing direction.

 

How do we determine the maximum stretch. I know how to do a) but i dont kno how to do b). The answer for a is 0.48 m/s. Can someone please help me.

Posted

When something is fully stretched just like a pendulum at the top of it's arc it's velocity is 0, knowing that you should be able to do pretty much the reverse of what you did for part a

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.