When a 0.106 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 5.20 cm, compressing the spring 1.70 cm.
To find the instantaneous acceleration of the mass when it is raised 5.20 cm and compresses the spring 1.70 cm, we can use Hooke's Law and Newton's second law of motion.
1. First, let's find the spring constant (k) using the given information. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, this can be represented as F = -kx, where F is the force, k is the spring constant, and x is the displacement.
In this case, we have a 0.106 kg mass suspended at rest, and the spring stretches 3.50 cm. As the mass is in equilibrium, the force exerted by the spring is equal to the weight of the mass. So, we have the equation:
k(0.0350 m) = mg
Rearranging the equation, we get:
k = (mg) / (0.0350 m)
2. Now that we have the spring constant, let's find the force acting on the mass at the new position. As the mass is raised 5.20 cm and compresses the spring 1.70 cm, the total displacement (x) from the equilibrium position is the sum of these two displacements:
x = 0.0520 m - (-0.0170 m)
x = 0.0690 m
Using Hooke's Law again, the force (F) acting on the mass can be calculated:
F = -kx
3. Finally, we can find the instantaneous acceleration (a) using Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be represented as F = ma, where F is the net force, m is the mass, and a is the acceleration.
In this case, the net force acting on the mass is the force exerted by the spring (F), and the mass is 0.106 kg. So, we have:
F = ma
Rearranging the equation, we get:
a = F / m
Substituting the value of F calculated from Hooke's Law and the given mass, we can obtain the acceleration (a).
Please note that the units used throughout the calculations should be consistent (e.g., meters and kilograms).