*I am lost on a bunch of different problems, and this is one of them. Please note that I am NOT asking for final answers, just some guidance so that I can figure out how to do the problem!!! I know that a lot of times on this site it is commented that we are fishing for answers, but I just need some guidance on some stuff I honestly cannot figure out how to do, no matter how easy it might seem to others.

A 28.0 kg child plays on a swing having support ropes that are 1.90 m long. A friend pulls her back until the ropes are 45.0 degrees from the vertical and releases her from rest.

What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?
Answer expressed in J.

How fast will she be moving at the bottom of the swing?
Answer expressed in m/s.

How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
Answer expressed in J.

Draw the figure. What is the height of the child? (I think it is 1.90-1.90cos45, check that)

PE = mgh
KE at bottom= 1/2 mv^2 which will equal the max PE at the top.

Work in tension? Do the ropes stretch? Work is only done by a force if it travels in the direction of the force.

To solve this problem, let's break it down into smaller parts.

1. First, let's find the potential energy for the child just as she is released compared to the potential energy at the bottom of the swing.

The potential energy of an object is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. At the initial position, the height is the vertical height of the swing, which is given by h = 1.90 - 1.90cos(45). Calculate this value to find the potential energy at the initial position.

Next, at the bottom of the swing, the potential energy is zero because the height is at its minimum. So, the potential energy for the child just as she is released compared to the potential energy at the bottom is the potential energy at the initial position.

2. Next, let's find the speed at the bottom of the swing.

The maximum kinetic energy of the child at the bottom of the swing is equal to the maximum potential energy at the top of the swing. This is because energy is conserved in a closed system, neglecting any energy losses due to friction or air resistance.

Therefore, set the maximum potential energy at the top equal to the maximum kinetic energy at the bottom and solve for the speed using the equation KE = 1/2 mv², where m is the mass and v is the speed.

3. Finally, let's find the work done by the tension in the ropes as the child swings from the initial position to the bottom.

The work done by a force is given by the equation W = Fd cos θ, where F is the force, d is the displacement, and θ is the angle between the force vector and the displacement vector. In this case, the tension in the ropes is the force, the displacement is the vertical height of the swing, and the angle is 45 degrees.

Calculate the work done by the tension using this equation.

Remember to substitute the given values into the equations and double-check your calculations to get the final answers.

I hope this guidance helps you figure out how to solve the problem step by step!

the tension is 0