University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

399

Pages:

5
Author:

Jabin F.

# Buoyant force on an object is weight of liquid displaced by the object. B = mg = V * ρ * g where: B = buoyant force m = mass of liquid displaced V = volume of object inside liquid ρ = density of liquid g = acceleration due to gravity Direction is always upward. In figure shown block is compressed against the spring, but is not attached to spring. What is max. height upto which block will rise one being released. Solution: Let P.E. of level by shown in figure when spring is compressed by a compression of x = 3mg/k. Then apply conservation of energy: (KE + PE)1 = (KE + PE)2 1/2 m² + 1/2 k² + mgx = 1/2 m² + 1/2 k² + mgx x = 0 x = 0 h = 3mg/k h = 0 Q. If the black is slightly displaced in x direction then write its equation of motion. Solution: Spring Force, F₃ = k(Je²+x²-1) EOM Fxnet = max-IF, Ca10 max = -2k (Je²+x²-ex max = -2kx(1-x) for slight displacement xxl. tends to x wrt I. ax = 0 thus we can't neglect 07 Note: #If we neglect x in comparison to l due to which we get zero then it is not acceptable result. #If approximation leads to zero value then apply binomial expansion. for x<< l, binomial expansion -9kx(1-(1-1/2)^1/2) = max => -9kx(1-(1-1/2)) = max using (1+x)^n = 1+nx+n(n-1)/2 x² + n(n-1)(n-2)/6 x³ + ... -9kx[1-(1-1/2)] = max => -kx³ = max => x = -kx³/m²