Problem 62
Question
A ball of mass \(1.84 \mathrm{~kg}\) is dropped from a height \(y_{1}=$$1.49 \mathrm{~m}\) and then bounces back up to a height of \(y_{2}=0.87 \mathrm{~m}\) How much mechanical energy is lost in the bounce? The effect of air resistance has been experimentally found to be negligible in this case, and you can ignore it.
Step-by-Step Solution
Verified Answer
Answer: The mechanical energy lost in the bounce is 11.06 J.
1Step 1: Calculate initial gravitational potential energy
Before the ball is dropped, its entire mechanical energy is in the form of gravitational potential energy. We can calculate this using the formula:
\(E_{p1} = m \cdot g \cdot y_{1}\)
where \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, and \(y_1\) is the initial height.
2Step 2: Calculate final gravitational potential energy
After the ball bounces, it reaches a height \(y_2\). At this point, the final gravitational potential energy can be calculated using the same formula:
\(E_{p2} = m \cdot g \cdot y_{2}\)
3Step 3: Calculate the mechanical energy lost
The mechanical energy lost during the bounce is the difference between the initial and final gravitational potential energies:
\(E_{lost} = E_{p1} - E_{p2}\)
Now, we just need to plug in the given values and calculate the energy lost.
4Step 4: Substitute the values and solve
Using the given values, we can calculate the energy lost:
- Mass of the ball, \(m = 1.84\) kg
- Initial height, \(y_{1} = 1.49\) m
- Final height, \(y_{2} = 0.87\) m
We also need the value of gravitational acceleration, \(g = 9.81 \mathrm{~m/s^2}\).
Now, we can calculate the initial and final gravitational potential energies:
\(E_{p1} = (1.84 \mathrm{~kg}) \cdot (9.81 \mathrm{~m/s^2}) \cdot (1.49 \mathrm{~m}) = 26.76 \mathrm{~J}\)
\(E_{p2} = (1.84 \mathrm{~kg}) \cdot (9.81 \mathrm{~m/s^2}) \cdot (0.87 \mathrm{~m}) = 15.70 \mathrm{~J}\)
Finally, we can find out the mechanical energy lost:
\(E_{lost} = 26.76 \mathrm{~J} - 15.70 \mathrm{~J} = 11.06 \mathrm{~J}\)
Therefore, the mechanical energy lost in the bounce is \(11.06 \mathrm{~J}\).
Key Concepts
Gravitational Potential EnergyConservation of EnergyElastic Collision
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. The higher an object is above the ground, the more potential energy it has, because there's more distance it can fall and more work that gravity can do on it.
To understand GPE better, let's look at the ball in our exercise. Before the ball is dropped, all its energy is stored as GPE, which can be calculated using the formula:
\(E_{p} = m \times g \times y\)
where \(m\) is the mass of the ball in kilograms, \(g\) is the acceleration due to gravity (approximately \(9.81 \text{m/s}^2\) on Earth), and \(y\) is the height above the ground in meters.
The concept of GPE is crucial in many physics problems, especially in understanding how energy is transferred from one form to another as objects move within a gravitational field.
To understand GPE better, let's look at the ball in our exercise. Before the ball is dropped, all its energy is stored as GPE, which can be calculated using the formula:
\(E_{p} = m \times g \times y\)
where \(m\) is the mass of the ball in kilograms, \(g\) is the acceleration due to gravity (approximately \(9.81 \text{m/s}^2\) on Earth), and \(y\) is the height above the ground in meters.
The concept of GPE is crucial in many physics problems, especially in understanding how energy is transferred from one form to another as objects move within a gravitational field.
Conservation of Energy
The principle of conservation of energy states that energy in a closed system cannot be created or destroyed; it can only be transformed from one form to another or transferred from one object to another. This means that the total energy of the system remains constant over time, as long as there's no external work done on it.
In our problem, the mechanical energy of the ball would remain the same if not for energy losses such as air resistance or deformation during the bounce. Since air resistance is negligible and not to be considered here, the mechanical energy loss we find is mainly due to the inelastic nature of the bounce, which is not a perfectly elastic collision. By calculating the initial and final GPE of the ball, we can deduce how much mechanical energy is lost in the process.
In our problem, the mechanical energy of the ball would remain the same if not for energy losses such as air resistance or deformation during the bounce. Since air resistance is negligible and not to be considered here, the mechanical energy loss we find is mainly due to the inelastic nature of the bounce, which is not a perfectly elastic collision. By calculating the initial and final GPE of the ball, we can deduce how much mechanical energy is lost in the process.
Elastic Collision
An elastic collision is a type of collision where the total kinetic energy and total momentum are conserved. In these collisions, objects bounce off each other without any loss in their overall mechanical energy. Examples include billiard balls hitting each other or a super-bouncy ball hitting a hard surface.
In contrast, the bounce of our ball is not perfectly elastic, as evidenced by the measured loss in GPE and, therefore, mechanical energy. Some of the ball's mechanical energy is transformed into other forms, like heat or sound, or is used to change the shape of the ball temporarily, all of which do not contribute to the ball's bounce-back height. This is why the ball only returns to a certain height lower than the original, showing a loss in mechanical energy. The study and understanding of elastic collisions are fundamental in physics as they help us to predict outcomes in systems where energy conservation is key.
In contrast, the bounce of our ball is not perfectly elastic, as evidenced by the measured loss in GPE and, therefore, mechanical energy. Some of the ball's mechanical energy is transformed into other forms, like heat or sound, or is used to change the shape of the ball temporarily, all of which do not contribute to the ball's bounce-back height. This is why the ball only returns to a certain height lower than the original, showing a loss in mechanical energy. The study and understanding of elastic collisions are fundamental in physics as they help us to predict outcomes in systems where energy conservation is key.
Other exercises in this chapter
Problem 57
A spring with a spring constant of \(500 . \mathrm{N} / \mathrm{m}\) is used to propel a 0.500 -kg mass up an inclined plane. The spring is compressed \(30.0 \m
View solution Problem 60
A 70.0 -kg skier moving horizontally at \(4.50 \mathrm{~m} / \mathrm{s}\) encounters a \(20.0^{\circ}\) incline. a) How far up the incline will the skier move b
View solution Problem 63
A car of mass \(987 \mathrm{~kg}\) is traveling on a horizontal segment of a freeway with a speed of \(64.5 \mathrm{mph}\). Suddenly, the driver has to hit the
View solution Problem 64
Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 10.0 -kg mass is released and falls through
View solution