# A 22.0 kg child plays on a swing having support ropes that are 2.00 m long. A friend pulls her back until the ropes are 45.0 ∘ 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? How fast will she be moving at the bottom of the swing? How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

## Well, well, well, let's swing into action and answer these questions!

First things first, let's tackle the potential energy. Just as the child is released, her potential energy will be at its maximum because she's at the highest point of the swing. As she swings to the bottom, her potential energy decreases and her kinetic energy increases. So, we can say that the potential energy at the top is equal to the kinetic energy at the bottom. It's like a little trade-off between potential and kinetic - one goes up, the other goes down.

Now, as for the speed at the bottom of the swing, we can use the conservation of energy principle. Assuming there's no energy loss due to friction or air resistance, we can equate the initial potential energy to the final kinetic energy. So, we can use the equation:

mgh = (1/2)mv^2

Where m is the mass of the child (22.0 kg), g is the acceleration due to gravity (9.8 m/s^2), h is the vertical distance from the top to the bottom (2.00 m), and v is the velocity at the bottom (what we're looking for). With a little bit of math, we can solve for v and find out the child's speed.

As for the work done by the tension in the ropes, it will be equal to the change in total mechanical energy. So, the work can be calculated by subtracting the potential energy at the top from the potential energy at the bottom. But remember, work is also equal to the change in kinetic energy. So, we can also calculate it by subtracting the initial kinetic energy (which is zero, as the child was at rest) from the final kinetic energy. Either way, the work done will be the same.

So, put on your physics cape and calculate away! Remember, humor is a swing-tastic way to make learning more fun!

## To find the potential energy of the child just as she is released, compared to the potential energy at the bottom of the swing, we need to use the concept of gravitational potential energy.

The formula for gravitational potential energy (PE) is given by:

PE = mgh

Where,

m = mass of the object (22.0 kg)

g = acceleration due to gravity (approximated as 9.8 m/s^2)

h = height above the reference point

At the initial position (when released), the height (h) can be calculated using trigonometry. Since the support ropes are 2.00 m long and they make an angle of 45.0 degrees from the vertical, we can use the sine function:

sin(45) = h / 2.00

Rearranging the equation, we find:

h = 2.00 * sin(45)

h ≈ 1.41 m

Now we can calculate the potential energy at the initial position:

PE(initial) = mgh

PE(initial) = 22.0 kg * 9.8 m/s^2 * 1.41 m

Next, let's calculate the potential energy at the bottom of the swing. At the bottom of the swing, the height (h) is 0, so the potential energy is also 0:

PE(bottom) = mgh

PE(bottom) = 22.0 kg * 9.8 m/s^2 * 0

The potential energy at the bottom of the swing is 0, as there is no height above the reference point.

To find the speed of the child at the bottom of the swing, we can use the principle of conservation of mechanical energy. At the highest point of the swing, the total mechanical energy (consisting of potential energy and kinetic energy) is equal to the mechanical energy at the lowest point of the swing.

At the highest point,

PE(initial) + KE(initial) = PE(bottom) + KE(bottom)

KE(initial) = PE(bottom)

Since the potential energy at the bottom is 0, the initial kinetic energy is equal to the potential energy at the bottom:

KE(initial) = mgh

KE(initial) = 22.0 kg * 9.8 m/s^2 * 0

Therefore, the initial kinetic energy is also 0.

At the bottom of the swing, all potential energy is converted into kinetic energy. We can find the kinetic energy at the bottom using the formula:

KE(bottom) = (1/2) * mv^2

Since the potential energy at the bottom is 0, all the potential energy is converted into kinetic energy, so:

KE(bottom) = PE(initial)

Using this equation, we can find the speed (v) at the bottom of the swing:

(1/2) * mv^2 = 22.0 kg * 9.8 m/s^2 * 1.41 m

Finally, let's calculate the work done by the tension in the ropes as the child swings from the initial position to the bottom. Work is given by the formula:

Work = force * distance * cosine(angle)

The tension in the ropes is equal to the weight of the child, which is given by:

Force = mg

The distance covered by the child from the initial position to the bottom is equal to the length of the rope, which is 2.00 m.

Work = mg * 2.00 m * cos(45)

Now you have all the steps to calculate the potential energy at the initial position, the speed at the bottom, and the work done by the tension in the ropes.

## To find the potential energy for the child just as she is released, we need to consider the change in potential energy. The potential energy at the bottom of the swing is given by the equation:

P.E. = mgh

Where:

m = mass of the child (22.0 kg)

g = acceleration due to gravity (approximately 9.8 m/s^2)

h = height from the bottom of the swing to the release point

At the bottom of the swing, the height is 0, so the potential energy is also 0. Therefore, the potential energy for the child just as she is released is equal to the potential energy at the bottom of the swing, which is 0.

Next, let's calculate the speed of the child at the bottom of the swing using the principle of conservation of energy. At the highest point of the swing, all potential energy is converted into kinetic energy.

Potential energy at the highest point (release point) = Kinetic energy at the lowest point

mgh = (1/2)mv^2

Where:

m = mass of the child (22.0 kg)

g = acceleration due to gravity (approximately 9.8 m/s^2)

h = height from the bottom of the swing to the release point

v = speed at the bottom of the swing (what we want to find)

Rearranging the equation, we get:

v^2 = 2gh

Substituting the given values:

v^2 = 2 * 9.8 m/s^2 * h

We know the length of the support ropes (2.00 m), and when the swing is released, the support ropes make an angle of 45.0° from the vertical. We can use trigonometry to find the height (h) from the bottom of the swing to the release point.

Using the trigonometric relationship:

sin(angle) = opposite/hypotenuse

sin(45°) = h/2

Solving for h, we have:

h = 2 * sin(45°) = 2 * 0.707 = 1.414 m

Substituting this value back into the equation for v^2:

v^2 = 2 * 9.8 m/s^2 * 1.414 m

v^2 = 27.97

So, the speed of the child at the bottom of the swing is approximately 5.29 m/s (taking the square root of 27.97).

Finally, let's calculate 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:

Work = Force * Distance * cos(angle)

In this case, the force is the tension in the ropes, the distance is the length of the circular arc (2πr), and the angle is the change in angle from the initial position to the bottom.

The tension in the ropes provides the necessary centripetal force to keep the child swinging in a circular path. The tension force is equal to the weight of the child (mg), and the distance is the circumference of the swing.

The circumference of the swing is given by:

C = 2 * π * r

Where:

r = length of the ropes (2.00 m)

Substituting the values, we get:

C = 2 * 3.14 * 2.00 = 12.56 m

Now we can calculate the work done by the tension force:

Work = (mg) * C * cos(angle)

Angle = 45°

Work = (22.0 kg * 9.8 m/s^2) * 12.56 m * cos(45°)

Work = 220.0 N * 12.56 m * 0.707

Work = 1956.32 Joules

So, the work done by the tension in the ropes as the child swings from the initial position to the bottom is approximately 1956.32 Joules.

## U = m g h = m g 2(1-cos 45)

(1/2) m v^2 = U

The tension is PERPENDICULAR to the motion, does ZERO work. Why did you ask !!! ???