Question: A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22

A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5 m/s. The block oscillates on the spring without friction.

  • 1) What is the spring constant of the spring?

  • 2 )What is the oscillation frequency?

  • 3) After t = 0.3 s what is the speed of the block?
  •  
  • 4) What is the magnitude of the maximum acceleration of the block?

  • 5) At t = 0.3 s what is the magnitude of the net force on the block?

  • 6) Where is the potential energy of the system the greatest?

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock

Lets solve these step by step 1 Spring Constant k Hookes Law states F kx where F is the force exerted by the spring k is the spring constant and x is ... View full answer

blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Physics Questions!