a 0.60kg block attached to a spring with a force constant 130n/m. is free to move on a frictionless, horizental surfce . The block is realesed from rest when the spring is stretched 0.13m. at the instant the block is relesed find force and accelration??/

F = 0.13m * 130N/m = 16.9 N.

a = F/M = 16.9/0.60 = 28.2 m/s^2.

To find the force and acceleration of the block, we can start by using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.

The formula for Hooke's Law is: F = kx,

Where:
F is the force exerted by the spring,
k is the force constant or spring constant (k = 130 N/m in this case),
x is the displacement of the spring from its equilibrium position.

In this case, the displacement of the spring is given as 0.13m. Substituting the values into the formula:

F = (130 N/m) * (0.13m)
F = 16.9 N

So, the force exerted by the spring is 16.9 N.

Next, to find the acceleration of the block, we can use 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.

The formula for Newton's second law is: F = ma,

Where:
F is the net force acting on the object,
m is the mass of the object (m = 0.60 kg in this case),
a is the acceleration of the object.

We have already found the force exerted by the spring as 16.9 N. Substituting the values into the formula:

16.9 N = (0.60 kg) * a

Solving for a:

a = 16.9 N / 0.60 kg
a = 28.2 m/s^2

So, the acceleration of the block is 28.2 m/s^2.