Question

A classmate throws a \(1.0-\mathrm{kg}\) book from a height of \(1.0 \mathrm{~m}\) above the ground straight up into the air. The book reaches a maximum height of \(3.0 \mathrm{~m}\) above the ground and begins to fall back. Assume that \(1.0 \mathrm{~m}\) above the ground is the reference level for zero gravitational potential energy. Determine a) the gravitational potential energy of the book when it hits the ground. b) the velocity of the book just before hitting the ground.

Short answer

Answer: The gravitational potential energy of the book when it hits the ground is 0 J, and its velocity just before hitting the ground is 6.26 m/s.

Step-by-step solution

Determine the change in gravitational potential energy

The book was initially thrown from a height of 1.0 m and reached a maximum height of 3.0 m before falling back to the ground. We are asked to find the gravitational potential energy of the book when it hits the ground. As we have taken 1.0 m above the ground as the reference level for zero gravitational potential energy, when the book hits the ground its height with respect to the reference level is 0. Therefore, the change in gravitational potential energy is given by PE = mgh = (1.0 kg) * (9.81 m/s²) * (0 m) = 0 J. So, the gravitational potential energy when it hits the ground is 0 J. a) Gravitational potential energy when it hits the ground: 0 J

Calculate the potential and kinetic energy at the highest point

We need to determine the potential and kinetic energy of the book at the highest point to find the total mechanical energy. The potential energy at the highest point is given by PE = mgh = (1.0 kg) * (9.81 m/s²) * (2.0 m) = 19.62 J (As the maximum height is 3.0 m, the difference in height from the reference level is 3.0 m - 1.0 m = 2.0 m). At the highest point, the book momentarily comes to a stop, so its kinetic energy is 0. Potential energy at the highest point: 19.62 J Kinetic energy at the highest point: 0 J

The Conservation of Mechanical Energy

When the book is just before hitting the ground, the potential energy is 0. Then, by conservation of mechanical energy, the kinetic energy becomes equal to the mechanical energy we found at the highest point. Hence, KE = 19.62 J. At the highest point: KE_high + PE_high = KE_ground + PE_ground

Determine the book's velocity just before hitting the ground

Now we can determine the book's velocity just before hitting the ground. We know the kinetic energy at this point is 19.62 J. Using the formula KE = (1/2)mv², we can solve for v. 19.62 J = (1/2)(1.0 kg)v² Rearrange the formula to get v²: v² = (2 * 19.62 J) / (1.0 kg) = 39.24 Now, take the square root to find the velocity: v = sqrt(39.24) = 6.26 m/s b) The book's velocity just before hitting the ground: 6.26 m/s

Key concepts

Conservation of Mechanical Energy

When we talk about the conservation of mechanical energy, we refer to an important principle in physics that states that in an isolated system, the total mechanical energy remains constant if only conservative forces are acting. Mechanical energy is the sum of kinetic and potential energy.
In our specific context, when the book is thrown into the air, it has gravitational potential energy because of its height, and kinetic energy due to its motion. As it rises, its kinetic energy decreases while its potential energy increases.
When it reaches the maximum height, the velocity momentarily reaches zero, and thus, the kinetic energy is zero, while the potential energy reaches its maximum value. This entire potential energy is then converted back into kinetic energy as the book falls back to the ground.

• Total Energy at any point = Kinetic Energy (KE) + Potential Energy (PE)
• Energy doesn’t "get lost"; it just changes forms

For the problem described, when the book was at the highest point, it had a potential energy of 19.62 J and a kinetic energy of 0 J. Just before hitting the ground, all this energy converts to kinetic energy (19.62 J), because at ground level the potential energy is zero. This showcases the conservation of mechanical energy.

Kinetic Energy Calculation

Kinetic energy is defined as the energy an object possesses due to its motion. It can be calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass and \( v \) is the velocity of the object.
In order to find how fast the book is moving just before it hits the ground, we used this concept. Since we already knew that the kinetic energy just before the impact was 19.62 J, the same as the total mechanical energy at the highest point, we used that to calculate velocity:

• Set the equation: \( 19.62 \text{ J} = \frac{1}{2}(1.0 \text{ kg})v^2 \)
• Solve for \( v^2 \): \( v^2 = \frac{2 \times 19.62 \text{ J}}{1.0 \text{ kg}} = 39.24 \)
• Take the square root to solve for \( v \): \( v = \sqrt{39.24} \approx 6.26 \text{ m/s} \)

So, the speed of the book right before it hits the ground is approximately 6.26 m/s. This calculation reflects how energy transforms during the projectile's motion.

Projectile Motion Analysis

Projectile motion occurs when an object is thrown into the air with an initial velocity and follows a curved path under the influence of gravity alone. In the given problem, our classmate throws the book straight up. This is a classic example of vertical projectile motion.
Let's break down what happens in projectile motion:

• The initial velocity makes the book ascend upward against the force of gravity.
• As the book rises, gravity causes its upward speed (and kinetic energy) to decrease until it reaches the peak, or the maximum height, at which point its velocity is instantaneously zero.
• No horizontal motion affects the book in this case, simplifying the analysis to a vertical up-down motion.
• Then, it descends, accelerating due to gravity, converting potential energy back into motion as kinetic energy returns.

The path is symmetrical in time and distance when air resistance is negligible. Understanding this motion helps in various applications, from engineering to sports science, to comprehend how energy conservation and forces interact in moving objects.