Spring force, find the speed at equilibrium

[P-Illiterate]

A .50 kg block sliding on horizontal frictionless surface is attached to one end of a horizontal spring (with k = 500 N/M) whose other end is fixed. The block has a kinetic energy of 20J as it pass through its equilibrium position (the point at which the spring force is zero.)

1. what is the speed of the block as it passes through its equilibrium position?
2. at what rate is the spring doing work (power) as the block passes through the equilibrium pont?
3. what is the maximum compression of the spring?
4. at what rate is the spring doing work on the block as the block passes through a point 5 cm away from the sprig's equilibrium position and being stretched?

The Attempt at a Solution

1.
I know this is wrong but
do I use K - 1/mv2
20 =.5*.50*v²
v -8.9 m/sI missed my lecture the day when his was taught, I don't have any idea how to solve any of this ... can you please help me??

 

[kuruman]

Yes that is what you use. The 20 J of energy is the total energy of the oscillator no matter where the mass is as it moves back and forth.

 

[P-Illiterate]

can you give me hints on the other three questions too

 

[kuruman]

For 2 and 4, what is an expression for power that you can use that involves the speed?
For 3, how would you express the total energy at maximum compression?

 

[rwisz]

Sure, I can provide some guidance on how to approach these questions. First, let's start with the concept of equilibrium. In this scenario, the block is sliding on a horizontal frictionless surface, and is attached to one end of a horizontal spring. At equilibrium, the net force on the block is zero, so the block is not accelerating. This means that the kinetic energy of the block must be equal to the potential energy stored in the spring.

1. To find the speed of the block at equilibrium, we can use the equation for the potential energy of a spring: PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement from equilibrium. Since the block has a kinetic energy of 20J at equilibrium, we can set the potential energy equal to this value and solve for x. This will give us the maximum displacement of the block from equilibrium. Then, we can use the equation for the speed of an object undergoing simple harmonic motion: v = ω * x, where ω is the angular frequency of the oscillation. We can find ω by using the equation ω = √(k/m), where m is the mass of the block. Plugging in the values, we can calculate the speed of the block at equilibrium.

2. The rate at which the spring is doing work (power) as the block passes through the equilibrium point can be found using the equation P = F * v, where F is the force exerted by the spring and v is the velocity of the block. At equilibrium, the spring force is zero, so the power will also be zero.

3. The maximum compression of the spring can be found by using the equation for the potential energy of a spring, PE = 1/2 * k * x^2, and setting it equal to the kinetic energy of the block at equilibrium. Solving for x will give us the maximum compression of the spring.

4. To find the rate at which the spring is doing work on the block as it passes through a point 5 cm away from equilibrium, we can use the same equation as before, P = F * v. However, this time the force exerted by the spring will not be zero, so we need to calculate the force first. We can use the equation F = -k * x, where x is the displacement from equilibrium. Plugging in the values, we can calculate the