Problem 119 - College Algebra with Modeling and Visualization

Question

Energy of a Falling Object \(\mathbf{A}\) ball with mass \(m\) is dropped from an initial height of \(h_{0}\) and lands with a final velocity of \(v_{f}\). The kinetic energy of the ball is \(K(v)=\frac{1}{2} m v^{2},\) where \(v\) is its velocity, and the potential energy of the ball is \(P(h)=m g h,\) where \(h\) is its height and \(g\) is a constant. (a) Show that \(P\left(h_{0}\right)=K\left(v_{f}\right) .\) (Hint: \(v_{f}=\sqrt{2 g h_{0}}\) ) (b) Interpret your result from part (a).

Step-by-Step Solution

Answer

The initial potential energy equals the final kinetic energy, demonstrating energy conservation.

Step 1: Understand Energy Conservation

When the ball is dropped, the potential energy at height \( h_0 \) is fully converted into kinetic energy just before it hits the ground. This is based on the conservation of mechanical energy, which states that the total energy remains constant if no external force is acting.

Step 2: Write the Expressions

The potential energy \( P(h_0) \) at the initial height is given by \( P(h_0) = mgh_0 \). The kinetic energy \( K(v_f) \) just before hitting the ground is \( K(v_f) = \frac{1}{2}mv_f^2 \).

Step 3: Substitute the Final Velocity

From the hint, we know \( v_f = \sqrt{2gh_0} \). Substitute \( v_f \) into the kinetic energy formula: \[ K(v_f) = \frac{1}{2}m(\sqrt{2gh_0})^2 \].

Step 4: Simplify the Kinetic Energy Expression

Simplify the equation \[ K(v_f) = \frac{1}{2}m(\sqrt{2gh_0})^2 \]. This becomes:\[ K(v_f) = \frac{1}{2}m(2gh_0) = mgh_0 \].

Step 5: Compare the Energies

Now compare \( P(h_0) = mgh_0 \) with \( K(v_f) = mgh_0 \). Since both expressions are equal, B) This confirms that the initial potential energy is equal to the kinetic energy just before impact.

Key Concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. For an object like a ball, its kinetic energy increases as it falls, because its velocity increases. The formula for kinetic energy is given by:

• \( K(v) = \frac{1}{2} m v^2 \)

Here, \( m \) represents the mass of the object, and \( v \) is its velocity. As the ball falls, gravity pulls it downwards, and potential energy is converted into kinetic energy. By the time the ball reaches the ground, its kinetic energy reaches the maximum because its speed is greatest at that point. This is a crucial aspect of the conservation of mechanical energy principle, where energy is not lost but transforms from one type to another.

Understanding kinetic energy helps us predict how fast an object will be moving when it reaches a given point, like the surface of the ground in this scenario. By knowing the initial conditions, such as the height from which the object falls, we can calculate the final velocity at which it impacts the ground. This is important for solving physics problems that involve moving bodies.

Potential Energy

Potential energy is stored energy, which is ready to be converted into another form, such as kinetic energy, when an object is in motion. The potential energy in this scenario is due to gravity and is dependent on the height of an object above the ground. The formula for potential energy is:

• \( P(h) = mgh \)

Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height from which the object is dropped. When the ball is at its initial height \( h_0 \), it has maximum potential energy and no kinetic energy.

As the ball begins to fall, its height decreases, causing its potential energy to decrease while its kinetic energy increases in equal measure. In an ideal system where no energy is lost to air resistance or other external factors, the potential energy at the start is completely converted into kinetic energy by the time the ball reaches the ground. This transformation is a perfect example of the conservation of mechanical energy in action.

Final Velocity

Final velocity, in this context, refers to the speed of the falling object just before it strikes the ground. This speed can be derived using energy conservation principles. When the ball is dropped, the potential energy it has is completely transformed into kinetic energy by the time it hits the ground. The expression given by:

• \( v_f = \sqrt{2gh_0} \)

allows us to calculate this final velocity because it links the initial height with the final speed, assuming no other forces interfere with the motion like air resistance.

In our problem, the hint provided, \( v_f = \sqrt{2gh_0} \), was essential to connect the given potential energy and the kinetic energy equations. Substituting this into the kinetic energy formula simplified the process, proving that \( mgh_0 = \frac{1}{2}mv_f^2 \) holds true. Therefore, final velocity is a crucial component that demonstrates how energy transformations occur when an object is in free fall. It also signifies the culmination of these energy changes right before the impact when all potential energy has been effectively transferred to kinetic energy.