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
To find the instantaneous acceleration of the mass when it is raised 5.70 cm and compresses the spring 2.20 cm, we can use Hooke's Law and Newton's Second Law.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be represented as:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
The relationship between force, mass, and acceleration is given by Newton's Second Law:
F = ma
Where F is the force, m is the mass, and a is the acceleration.
1. Find the spring constant:
Given that a 0.116 kg mass stretches the spring 3.50 cm, we can calculate the spring constant using Hooke's Law. Rearranging the formula, we have:
k = -F / x
Since the mass is in equilibrium, the force exerted by the spring is equal in magnitude but opposite in direction to the force due to gravity, which is given by:
F = mg
Substituting the values, we get:
k = -(mg) / x
k = - (0.116 kg * 9.8 m/s^2) / (0.035 m)
k ≈ -33.377 N/m
2. Find the acceleration when the mass is raised 5.70 cm:
When the mass is raised, it moves further from the equilibrium position. The displacement can be expressed as:
x = -0.057 m (negative because it is raised)
Using Hooke's Law, we can find the force exerted by the spring:
F = -kx
F = -(-33.377 N/m)(-0.057 m)
F ≈ 1.907 N
Now, we can use Newton's Second Law to find the acceleration:
F = ma
1.907 N = (0.116 kg) * a
a ≈ 16.43 m/s^2
Therefore, the instantaneous acceleration of the mass when it is raised 5.70 cm is approximately 16.43 m/s^2.