Problem 16 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 72 -kg trampoline artist jumps vertically upward from the top of a platform with a \(\begin{array}{llll}\text { speed of } & 4.5 \mathrm{~m} / \mathrm{s} . & (a) & \text { How }\end{array}\) fast is he going as he lands on the trampoline, \(2.0 \mathrm{~m}\) below (Fig. \(8-31\) )? (b) If the trampoline behaves like a spring of spring constant \(5.8 \times 10^{4} \mathrm{~N} / \mathrm{m}\), how far does he depress it?

Step-by-Step Solution

Answer

(a) 7.71 m/s; (b) 0.272 m.

Step 1: Determine the velocity using energy conservation

For part (a), we'll use the conservation of mechanical energy. As the artist moves from the platform to the trampoline, the total mechanical energy is conserved. So:\[ E_{initial} = E_{final} \]Where:- Initial energy \( E_{initial} = KE_{platform} + PE_{platform} \)- Final energy \( E_{final} = KE_{trampoline} + PE_{trampoline} \)Given that at the trampoline \( PE_{trampoline} = 0 \) (ground level), the equation becomes:\[ \frac{1}{2}mv_{i}^2 + mgh = \frac{1}{2}mv_{f}^2 \]Substitute the given values:\[ \frac{1}{2}(72)(4.5)^2 + 72(9.8)(2) = \frac{1}{2}(72)v_{f}^2 \]

Step 2: Solve for final velocity

Simplify and solve for the final velocity \( v_f \):\[ \frac{1}{2}(72)(20.25) + 72(19.6) = \frac{1}{2}(72)v_{f}^2 \]\[ 729 + 1411.2 = \frac{1}{2}(72)v_{f}^2 \]\[ 2140.2 = 36v_{f}^2 \]Divide both sides by 36:\[ v_{f}^2 = \frac{2140.2}{36} \approx 59.45 \]\[ v_{f} = \sqrt{59.45} \approx 7.71 \text{ m/s} \]

Step 3: Consider the harmonic motion for spring compression

For part (b), we need to determine how much the trampoline (acting as a spring) is compressed. Using the energy conservation principle again, the kinetic energy of the artist at the trampoline's surface is converted into spring potential energy:\[ \frac{1}{2}mv_{f}^2 = \frac{1}{2}kx^2 \]Where:- \( m = 72 \text{ kg} \)- \( v_f = 7.71 \text{ m/s} \) (velocity from part a)- \( k = 5.8 \times 10^4 \text{ N/m} \)Substitute the known values:\[ \frac{1}{2}(72)(7.71)^2 = \frac{1}{2}(5.8 \times 10^4)x^2 \]

Step 4: Solve for spring compression

Continuing from Step 3:\[ (72)(59.45) = (5.8 \times 10^4)x^2 \]\[ 4280.4 = 5.8 \times 10^4 x^2 \]Divide both sides by \( 5.8 \times 10^4 \):\[ x^2 = \frac{4280.4}{5.8 \times 10^4} \approx 0.0738 \]\[ x = \sqrt{0.0738} \approx 0.272 \text{ m} \]

Step 5: Result Summary

The final velocity of the artist as he lands on the trampoline is approximately \( 7.71 \text{ m/s} \), and he depresses the trampoline by approximately \( 0.272 \text{ m} \).

Key Concepts

Kinetic Energy

Kinetic energy is a form of energy that an object possesses due to its motion. It is expressed through the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) stands for the mass of the object and \( v \) represents its velocity.
This formula indicates that kinetic energy depends on both the mass and the square of the velocity of an object. Therefore, when the velocity of an object increases, its kinetic energy increases quadratically, meaning it grows even faster than linear growth. For example:

• If you double the velocity of an object while keeping its mass constant, the kinetic energy becomes four times greater.
• Similarly, if you triple the velocity, the kinetic energy increases by a factor of nine.

In the case of the trampoline artist, as he descends from the platform to the trampoline, his kinetic energy increases due to the acceleration caused by gravity. Starting with an initial speed of \( 4.5 \, \text{m/s} \), his velocity increases until he hits the trampoline. At this point, his kinetic energy is transformed into potential energy, as explained in the following sections.

Potential Energy

Potential energy is the energy stored in an object due to its position or state. In the context of gravitational potential energy, it is determined by the height of the object above ground level. The formula for gravitational potential energy is \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth), and \( h \) represents the height.
Potential energy is fundamentally different from kinetic energy because it depends on an object's position rather than its motion. In this scenario:

• The trampoline artist has potential energy while he is on the platform because he is above ground level.
• As he descends to the trampoline, this potential energy gets converted into kinetic energy.

By the time he reaches the trampoline, all his initial potential energy has transformed into kinetic energy. This transition showcases the law of conservation of energy, wherein energy changes from one form to another but remains constant in total amount.

Spring Compression

When the trampoline artist lands on the trampoline, it behaves like a spring. This principle involves transforming kinetic energy into the potential energy stored within the spring during compression. A spring's potential energy can be calculated using the equation: \( PE_{spring} = \frac{1}{2}kx^2 \), where \( k \) is the spring constant, and \( x \) is the amount of compression from its equilibrium position.
The spring constant \( k \) measures how stiff the spring is, while \( x \) reflects how much the spring is compressed. In this exercise:

• The artist's kinetic energy at the point of contact compresses the trampoline.
• This energy is stored in the compressed spring as potential energy until it releases and returns to equilibrium.

Ultimately, by solving the equation \( \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \), we find the compression \( x \), showing how much the trampoline depresses when the artist lands on it. Here, the trampoline compresses approximately \( 0.272 \, \text{m} \), illustrating the conversion of kinetic energy into elastic potential energy, governed by the spring constant's stiffness.