The potetial energy of a spring is 0.25 J when it is compressed 0.05m to the left of the equilibrium position. At this point it comes to its leftmost position. What is its kinetic energy when it returns to its equilibrium point?

a. zero
b. 0.25 J and the object is moving to the right.
c. -0.25 J and the object is moving to the left.
d. 0.25 J and the object is at rest.

Please help?

compressed to left, then it moves to right, KE=.25J

To solve this problem, we need to understand the relationship between potential energy and kinetic energy for a spring.

The potential energy of a spring can be calculated using the formula:

PE = 0.5kx^2

Where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

Given that the potential energy of the spring is 0.25 J when it is compressed 0.05m to the left of the equilibrium position, we can use this information to find the value of k.

0.25 J = 0.5k(0.05m)^2

k(0.0025m^2) = 0.25 J

k = 0.25 J / 0.0025m^2

k = 100 J/m

Now, let's consider what happens when the spring returns to its equilibrium point. At this point, the potential energy of the spring is zero because the spring is neither compressed nor stretched.

Since potential energy is converted into kinetic energy as the spring returns to its equilibrium position, the kinetic energy must be equal to the initial potential energy of the spring, which is 0.25 J.

Therefore, the correct answer is:

b. 0.25 J and the object is moving to the right.

To determine the kinetic energy of the object when it returns to its equilibrium point, we first need to understand the concept of energy conservation. Energy cannot be created or destroyed, so the total mechanical energy of the system remains constant.

In this case, the total mechanical energy of the system is the sum of the potential energy (PE) and the kinetic energy (KE) of the object. When the object is at its leftmost position, all of its potential energy is converted to kinetic energy.

Given that the potential energy of the spring is 0.25 J when it is compressed 0.05 m to the left, we can use the spring potential energy formula:

PE = (1/2) * k * x^2

Where:
PE is the potential energy (0.25 J),
k is the spring constant, and
x is the displacement from the equilibrium position (-0.05 m, since it is compressed to the left).

Plugging in the given values, we have:

0.25 J = (1/2) * k * (-0.05 m)^2

We can solve this equation to find the spring constant (k). Dividing both sides by 1/2 and (-0.05 m)^2, we get:

k = (0.25 J) / (1/2 * (-0.05 m)^2)
= 10 J/m^2

Now, at the equilibrium position, all potential energy has been converted to kinetic energy. Therefore, the kinetic energy (KE) of the object when it returns to its equilibrium point is also 0.25 J. Hence, the answer is:

b. 0.25 J and the object is moving to the right.