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
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
1 answer
1 answer