Question
A block of mass m = 2.00 kg is attached to a spring of force constant k = 5.00 x102 N/m
that lies on a horizontal frictionless surface. The block is pulled to a position xi = 5.00 cm
to the right of equilibrium and released from rest. Find
a) the work required to stretch the spring and
b) the speed the block has as it passes through equilibrium
that lies on a horizontal frictionless surface. The block is pulled to a position xi = 5.00 cm
to the right of equilibrium and released from rest. Find
a) the work required to stretch the spring and
b) the speed the block has as it passes through equilibrium
Answers
k = 5 * 10^2 Newtons / meter
x = 5 * 10^-2 meters
F = k x = 25 * 10^0 = 25 N
work in = integral k dx = (1/2) k x^2
= (1/2) * 5*10^2 * 25*10^-4 = (125/2) * 10^-2
= 1.25/2 = 0.625 Joules
then
(1/2) m v^2 = kinetic energy = 0.625 Joules
x = 5 * 10^-2 meters
F = k x = 25 * 10^0 = 25 N
work in = integral k dx = (1/2) k x^2
= (1/2) * 5*10^2 * 25*10^-4 = (125/2) * 10^-2
= 1.25/2 = 0.625 Joules
then
(1/2) m v^2 = kinetic energy = 0.625 Joules