Question

A 350 g block is attached to a vertical spring whose stiffness constant is 12 N/m . The block is released at the position where the spring is unextended.

a) What is the maximum extension of the spring?


b)How long does it take the block to reach the lowest point?

Answer 1

Fspring = -kx where x is extension up
= -12 x N

m = 0.35 kg

Weight = .35 * 9.8 = 3.45 N

How far will it drop before the force up = the force down? (the equilibrium position)
-k (-x) + m g = 0
12 x = -3.45
x = -.2875
How much potential energy did it lose going from x = 0 to x = -.2875 ?
m g h = 3.45 * .2875 = .992 Joules
that is kinetic energy as it passes the equilibrium point where x = -.2875
(1/2) m v^2 = .992
(1/2) .35 v^2 = .992
v = 2.38 m/s as it passes through equilibrium point
The total energy in the spring increases when it stops at the bottom to start back up> That increase is
.992 J
so the total energy stored in the spring is (1/2) 12 (.2875^2)+ .992
= 1.488 Joules
(1/2)(12) (total extension)^2 = 1.488
total extension = .4980 meters

amplitude of motion = .4980 - .2875 = .2105 meters

b) (1/2) period
where
period = 2 pi sqrt (M/k)
T = 2 pi sqrt(.35/12) = 1.07 seconds
so
0.5 seconds