A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.10 N is applied. A 0.440-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (ASSUME THAT THE DIRECTION OF THE INITIAL DISPLACEMENT IS POSITIVE.)
Determine the velocity and acceleration of the particle when t = 0.500 s. (Indicate the direction with the sign of your answer.)
To determine the velocity and acceleration of the particle when t = 0.500 s, we can use the concepts of energy conservation and simple harmonic motion.
First, let's calculate the potential energy stored in the spring when it is stretched. The potential energy in a spring is given by the equation:
PE = (1/2) k x^2
Where PE is the potential energy, k is the spring constant, and x is the displacement from the relaxed length.
Given that the spring stretches 3.00 cm and a force of 7.10 N is applied, we can find the spring constant using Hooke's Law:
F = k x
Solving for k:
k = F / x
Let's plug in the values:
k = 7.10 N / 0.03 m = 236.67 N/m
Now, let's calculate the potential energy:
PE = (1/2) (236.67 N/m) (0.03 m)^2 = 0.1 J
Since the particle is released from rest, all the potential energy will be converted into kinetic energy when it reaches x = 5.00 cm.
The equation that relates the potential energy to the kinetic energy is:
KE = (1/2) m v^2
Where KE is the kinetic energy, m is the mass of the particle, and v is the velocity.
Let's rearrange this equation to solve for v:
v = sqrt((2 KE) / m)
Substituting the known values into the equation:
KE = 0.1 J
m = 0.440 kg
v = sqrt((2 * 0.1 J) / 0.440 kg) = sqrt(0.454545...) m/s ≈ 0.674 m/s
So, the velocity of the particle at t = 0.500 s is approximately 0.674 m/s in the positive direction.
To find the acceleration at t = 0.500 s, we can use the equation for acceleration in simple harmonic motion:
a = -ω^2 x
Where a is the acceleration, ω (omega) is the angular frequency, and x is the displacement from the equilibrium position.
The angular frequency is given by:
ω = sqrt(k / m)
Substituting the known values:
ω = sqrt((236.67 N/m) / 0.440 kg) = sqrt(538.75) rad/s ≈ 23.20 rad/s
The displacement at t = 0.500 s is x = 5.00 cm = 0.05 m. Therefore,
a = - (23.20 rad/s)^2 * 0.05 m ≈ -26.88 m/s^2
The acceleration at t = 0.500 s is approximately -26.88 m/s^2 in the negative direction.
So, the velocity of the particle at t = 0.500 s is approximately 0.674 m/s in the positive direction, and the acceleration is approximately -26.88 m/s^2 in the negative direction.