A 500 g block is attached to a spring with a spring constant of 250 N/m. The block is pulled back a distance of 10 cm from its equilibrium position and released. Determine the following:
a) The potential energy stored in the spring when the block is pulled back. b) The maximum acceleration of the block. c) The maximum speed of the block.
a) The potential energy stored in the spring when the block is pulled back can be calculated using Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position.
The potential energy stored in a spring can be given by the equation:
PEspring = (1/2) * k * x^2
where PEspring is the potential energy stored in the spring, k is the spring constant, and x is the displacement from the equilibrium position.
Given:
Substituting the given values into the equation, we get:
PEspring = (1/2) * 250 N/m * (0.10 m)^2
Solving the equation:
PEspring = 1.25 J
Therefore, the potential energy stored in the spring when the block is pulled back is 1.25 Joules.
b) The maximum acceleration of the block can be calculated using Newton's second law, which states that the force acting on an object is equal to the product of its mass and acceleration.
The force exerted by the spring can be given by Hooke's Law:
Fspring = -k * x
where Fspring is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
We can equate this force to the mass of the block times its acceleration:
-k * x = m * a
Rearranging the equation to solve for acceleration:
a = (-k * x) / m
Substituting the given values:
a = (-250 N/m * 0.10 m) / 0.5 kg
Solving the equation:
a = -5 m/s^2
Since acceleration is a vector quantity, the negative sign indicates that the block's acceleration is in the opposite direction of the displacement.
Therefore, the maximum acceleration of the block is 5 m/s^2.
c) The maximum speed of the block can be calculated using the equation for simple harmonic motion, which relates the maximum speed of an object to the amplitude and angular frequency.
The angular frequency, ω, can be calculated using the equation:
ω = sqrt(k / m)
Given:
Substituting the given values into the equation, we get:
ω = sqrt(250 N/m / 0.5 kg)
Simplifying the equation:
ω = sqrt(500 N/kg)
ω = sqrt(500) rad/s
The maximum speed, vmax, can be calculated using the equation:
vmax = ω * A
where A is the amplitude, which is equal to the displacement from the equilibrium position.
Given:
Substituting the given values into the equation, we get:
vmax = sqrt(500) rad/s * 0.10 m
Simplifying the equation:
vmax = 7.07 m/s
Therefore, the maximum speed of the block is 7.07 m/s.