# samp11 - Mg – kx – kx = – kx . Thus, the spring acts...

This preview shows page 1. Sign up to view the full content.

SAMPLE PROBLEM (Class 11) ANS. a) When the block of mass M = 2.00 kg is hung from the spring, the spring stretches to a new equilibrium position, x 0 , at which the net force on the block is zero. That is, Mg – kx 0 = 0. Then, if the spring is stretched or compressed an additional distance, x , the net force exerted will be, F = Mg – k ( x 0 + x ) =
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Mg – kx – kx = – kx . Thus, the spring acts the same as it would when displaced from the unstretched position. The spring stretches until the force on the m = 300 g body is zero. F = mg – kx = 0, so k = mg/x = (0.300 kg)(9.8 m/s²)/0.0200 m) k = 147 N/m m / N 147 kg 00 . 2 2 2 b) π = = k M T T = 0.733 s...
View Full Document

## This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online