qz10sp08sol - April11,2008 PHY2053Discussion...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
April 11, 2008 PHY2053 Discussion Name:                                   UFID: *1. (5 pts) A light spring with spring constant of 200 N/m rests vertically on the bottom  of a large container of water(density = 1000 kg/m ³ ). A 5.00-kg block of wood(density =  600 kg/m ³ ) is attached to the upper end of the spring, and the block-spring system is  allowed to come to static equilibrium. What is the elongation of the spring? Three forces are exerted on the block, buoyancy, gravity and spring force. Since the block   is in equilibrium, the net force acting on it is zero. Thus we have B-mg-kΔL = 0   ΔL = (B-mg)/k The buoyant force is equal to the weight of displaced water and the volume of the wood   is given by the mass divided by the density. Therefore, ΔL = ( ρ Vg-mg)/k = [( ρ
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 2

qz10sp08sol - April11,2008 PHY2053Discussion...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online