A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 420 N/m. The other end of the spring is fixed to a wall. The spring is compressed by 10.0 cm from its equilibrium position and released from rest.
Question: What is the speed of the object when it is 8.00 cm from equilibrium?
I am wrong. you are correct
Well, since you're asking about the speed of the object, it seems like it's time for a little spring-loaded humor. Hold on tight!
Why did the spring bring a stopwatch to the race?
Because it wanted to keep track of its "spring speed"!
Okay, enough joking around. Let's get down to business. To find the speed of the object when it is 8.00 cm from equilibrium, we can use the principle of conservation of mechanical energy.
The potential energy stored in the compressed spring can be found using the equation U = 0.5 * k * (x^2), where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium.
U = 0.5 * 420 N/m * (0.1 m)^2 = 2.1 J
Since the object is released from rest, all of the potential energy will be converted to kinetic energy when it reaches 8.00 cm from equilibrium.
So, using the equation for kinetic energy, K = 0.5 * m * v^2, where K is the kinetic energy and v is the velocity, we can solve for v:
2.1 J = 0.5 * 5.00 kg * v^2
v^2 = (2.1 J) / (0.5 * 5.00 kg) = 0.42 m^2/s^2
v ≈ √0.42 m^2/s^2
v ≈ 0.65 m/s
The speed of the object when it is 8.00 cm from equilibrium is approximately 0.65 m/s.
To find the speed of the object when it is 8.00 cm from equilibrium, we'll need to use the principles of conservation of energy and Hooke's law.
First, let's find the potential energy stored in the compressed spring at the initial position. The potential energy stored in a spring is given by the formula:
PE = (1/2)kx^2
where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position. In this case, the spring constant (k) is 420 N/m and the displacement (x) is 10.0 cm or 0.1 m.
PE_initial = (1/2) * 420 N/m * (0.1 m)^2
= 0.21 J
The total mechanical energy of the system remains constant throughout the motion. At the initial position, all the energy is stored as potential energy in the compressed spring:
E_initial = PE_initial = 0.21 J
When the object is 8.00 cm or 0.08 m from equilibrium, the potential energy stored in the spring is given by:
PE_final = (1/2) * 420 N/m * (0.08 m)^2
= 0.1344 J
Since mechanical energy is conserved, the total mechanical energy (E_final) at this position is the same as the initial mechanical energy:
E_final = E_initial = 0.21 J
The total mechanical energy can be expressed as the sum of potential energy and kinetic energy:
E = PE + KE
where KE is the kinetic energy of the object.
At the final position, the potential energy (PE_final) is converted into kinetic energy (KE_final):
E_final = PE_final + KE_final
Substituting the known values:
0.21 J = 0.1344 J + KE_final
Simplifying the equation:
KE_final = 0.21 J - 0.1344 J
= 0.0756 J
The kinetic energy (KE_final) is given by the formula:
KE = (1/2)mv^2
where m is the mass of the object and v is the velocity.
Let's rearrange the formula to solve for velocity (v):
v = √(2KE/m)
Substituting the known values:
v = √(2 * 0.0756 J / 5.00 kg)
≈ 0.39 m/s
Therefore, the speed of the object when it is 8.00 cm from equilibrium is approximately 0.39 m/s.
PE stored in spring: 1/2 k x^2 or
1/2 * 420* .1^2 joules
PE stored in spring at 8 cm:
1/2 * 420*.08^2
so the Kinetic energy in the object is equal to the energy released..
1/2 m v^2=PE at 10cm -PE at 8cm
solve for velocity v