A 0.46 kg mass is attached to a spring with a spring constant of 145 N/m so that the mass is allowed to move on a horizontal frictionless surface. The mass is released from rest when the spring is compressed 0.14 m. Find the acceleration of the mass at the instant the spring is released. Answer in units of m/s2.
nvm. got it
To find the acceleration of the mass at the instant the spring is released, we can use Hooke's law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
In this case, the force exerted by the spring is given by F = -kx, where F is the force, k is the spring constant, and x is the displacement.
Given that the spring constant is 145 N/m, and the displacement is 0.14 m, we can calculate the force exerted by the spring:
F = -kx
F = -(145 N/m)(0.14 m)
F = -20.3 N
Since the force exerted by the spring is directed opposite to the displacement of the mass, we need to use the negative sign in the equation to indicate the direction.
Next, we can use Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. The net force in this case is the force exerted by the spring, and the mass is 0.46 kg.
a = F/m
a = (-20.3 N) / (0.46 kg)
a ≈ -44.13 m/s²
Therefore, the acceleration of the mass at the instant the spring is released is approximately -44.13 m/s².