A 0.500kg mass is suspended from a spring. A force of 1.00N stretches it an additional 35.00mm.?

Question

The force constant of the spring is 28.6 kg/s^2, and the period of motion which results when the mass is released is 23.8s.

What would the maximum acceleration and maximum velocity of the system be?

Answer 1

for the spring:
F = -kx

F = -28.6kg/s^2 * 0.035m
F = 1.0N

max acceleration:
F = ma => a = F/m
a = 1.0N/0.5kg = 2.0m/s^2

CORRECTION (Kirchwey pointed it out)
max velocity using Hooke's Law:
PE = 1/2*k*x^2
PE = 0.5*(28.6kg/s^2)*(0.035m)^2
PE = 0.0175175Nm

PE = 1/2mv^2 => v = sqrt(2*(0.0175175Nm)/0.5kg)
v = sqrt(0.07007)m/s
v= 0.265m/s

Answer 2

1st answer erroneously set force = energy (.5*m*v^2) and this resulted in velocity v being the square root of f/m or acceleration. I assume the question asks about the acceleration and velocity when the 1 N force is released and the mass oscillates on the spring. The radian frequency omega can be obtained two ways. As 2*pi/T it is 0.264 rad/s. However, this conflicts with the omega obtained from the spring and mass as sqrt(k/m), which is 7.559 rad/s. They would be closer if the mass were 500kg (0.239 rad/s) but still not in agreement.
First I'll assume the omega value of .264 derived from the period. Then the maximum velocity = amplitude * omega = .035 * .264 = 0.00924 m/s, and the acceleration = amplitude * omega^2 = 0.002439 m/s^2.
Next assuming omega = 7.559, velocity = .2646 and acceleration = 2 exactly, which checks with f/m.
You will need to resolve the conflict in the problem's data to decide which of these two answers is correct.

Answer 3

d = 0.04 m m = 0.fifty 5 kg mg = (0.fifty 5)(9.8) = 5.39 N stress appearing on spring = F = mg = 5.39 N stretch (deformation) of spring = d = 0.04 m Hooke's regulation of elastic forces: F = (ok)(d) {ok = stress consistent of spring} ok = F/d = 5.39/0.04 = _________ N/m ANS