When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm.

m•g=k•x₁

k= m•g/x₁=0.116•9.8/0.035= 32.5 N/m
When the spring rises 5.7 cm =0.057 m and the spring is compressed to x₂=0.022 m
F= k•x₂ =ma
a =k•x₂ /m=32.5•0.022/0.116 =6.16 m/s²

.116 * 9.81 = k (.035)

so
k = 32.5 N/m

F = .116*9.18+32.5*.022 = 1.14 + .715 = 1.855 N

1.855 = .116 a
a = 16 m/s^2

another way

spring is compressed .057 from equilibrium hanging

F = k (.057) = 32.5 * .057 = 1.85 N
a = 1.85/.116 = 16 m/s^2 again

To find the instantaneous acceleration of the mass, we can use Hooke's Law and the principles of simple harmonic motion.

Firstly, let's calculate the spring constant (k) using Hooke's Law equation:

F = -kx

Where:
F is the force applied to the spring,
k is the spring constant,
x is the displacement from the equilibrium position.

We know that when the mass is suspended at rest, the force applied to the spring is equal to the gravitational force acting on the mass, given by:

F = mg

Where:
m is the mass,
g is the acceleration due to gravity.

In this case, the mass is 0.116 kg, so the gravitational force is F = (0.116 kg)(9.8 m/s²) = 1.1368 N.

Since the mass is suspended at rest, the displacement (x) is the elongation of the spring, which is 3.50 cm or 0.035 m.

Plugging these values into Hooke's Law equation, we can solve for the spring constant (k):

1.1368 N = -k(0.035 m)

Rearranging the equation, we find:

k = -1.1368 N / -0.035 m = 32.53 N/m.

Now that we know the spring constant, we can find the angular frequency (ω) of the mass-spring system using the formula:

ω = √(k / m)

Where:
ω is the angular frequency,
k is the spring constant,
m is the mass.

Plugging in the values, we get:

ω = √(32.53 N/m / 0.116 kg) ≈ 5.913 rad/s.

Next, we can find the displacement of the mass when it is raised 5.70 cm or 0.057 m. This new displacement (x') can be calculated as the initial displacement (x) plus the compression of the spring (2.20 cm or 0.022 m):

x' = x + compressive displacement = 0.035 m + (-0.022 m) = 0.013 m.

To find the instantaneous acceleration (a) of the mass, we can use the equation for simple harmonic motion:

a = -ω²x'

Where:
a is the acceleration,
ω is the angular frequency,
x' is the displacement.

Plugging in the values, we have:

a = - (5.913 rad/s)²(0.013 m) ≈ -0.453 m/s².

Therefore, the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm, is approximately -0.453 m/s².