Problem 43 - University Physics with Modern Physics

Question

The froghopper (\(Philaenus spumarius\)), the champion leaper of the insect world, has a mass of 12.3 mg and leaves the ground (in the most energetic jumps) at 4.0 m/s from a vertical start. The jump itself lasts a mere 1.0 ms before the insect is clear of the ground. Assuming constant acceleration, (a) draw a free-body diagram of this mighty leaper during the jump; (b) find the force that the ground exerts on the froghopper during the jump; and (c) express the force in part (b) in terms of the froghopper's weight.

Step-by-Step Solution

Answer

The force exerted by the ground is approximately 49.3 N, around 408.75 times the froghopper's weight.

Step 1: Understanding the scenario

A froghopper has a mass of 12.3 mg and jumps vertically with an initial velocity of 4.0 m/s. The jump lasts 1.0 ms, requiring us to calculate the force exerted on the froghopper by the ground using constant acceleration.

Step 1: Draw a Free-Body Diagram

During the jump, the froghopper experiences two forces: the upward normal force exerted by the ground (\( F_{N} \)) and the downward gravitational force (weight, \( F_{G} \)). These forces can be visualized with arrows, \( F_{N} \) pointing upwards and \( F_{G} = mg \) pointing downwards, where \( m \) is the mass of the froghopper and \( g \) is the acceleration due to gravity.

Step 2: Calculating the acceleration

Using the kinematic equation \( v = u + at \), where \( v \) is the final velocity (4.0 m/s), \( u \) is the initial velocity (0 m/s), \( a \) is the acceleration, and \( t \) is the time of the jump (1.0 ms or 0.001 s), we solve for \( a \):\[a = \frac{v - u}{t} = \frac{4.0 \, \text{m/s}}{0.001 \, \text{s}} = 4000 \, \text{m/s}^2.\]

Step 3: Using Newton's second law

Apply Newton's second law to find the net force (\( F_{net} = ma \)), considering \( F_{net} = F_{N} - F_{G} \):\[F_{net} = ma = 12.3 \, \text{mg} \times 4000 \, \text{m/s}^2 = 0.0123 \, \text{kg} \times 4000 \, \text{m/s}^2 = 49.2 \, \text{N}.\]

Step 4: Calculate normal force

The normal force \( F_{N} \) must balance gravity and provide the net force for upward acceleration:\[F_{N} = F_{net} + F_{G} = 49.2 \, \text{N} + 0.0123 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 49.319 \, \text{N}.\]

Step 5: Expressing force in terms of weight

The weight of the froghopper is \( F_{G} = mg = 0.0123 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.120663 \, \text{N} \). We express the ground force in terms of weight:\[\frac{F_{N}}{F_{G}} = \frac{49.319 \, \text{N}}{0.120663 \, \text{N}} \approx 408.75.\]So, the ground force is approximately 408.75 times the froghopper's weight.

Key Concepts

Newton's Laws of Motion

Newton's Laws of Motion, formulated by Sir Isaac Newton, provide the foundation for understanding how objects move and interact in our world. The second law is particularly important for analyzing the movement of the froghopper. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This is expressed mathematically as \( F_{net} = ma \).

In this froghopper jump scenario, the frog's net force during the jump arises from the ground exerting an upward normal force, while gravity pulls it downward. Understanding these forces allows us to apply the second law to deduce how powerful its leap is. By calculating the net force and considering both the normal force and gravitational force, we can see exactly how impressive this small insect's jump is in comparison to its weight.

Kinematics

Kinematics focuses on the movement of objects without considering the forces that cause this movement. It allows us to predict how an object moves by using equations that relate displacement, velocity, acceleration, and time.

In the froghopper exercise, kinematics helps to determine the acceleration required to achieve the jump. We use the formula \( v = u + at \) to link the initial velocity, final velocity, acceleration, and time. Here, the froghopper's final velocity is 4.0 m/s, starting from rest (0 m/s) in a mere 1.0 ms (0.001 s). Solving the equation gives us an impressive acceleration of 4000 m/s², demonstrating the rapidity of the froghopper’s launch.

Free-Body Diagram

A free-body diagram is a simple visual tool used to illustrate the forces acting on an object. For the froghopper, the primary forces include the upward normal force (\( F_{N} \)) exerted by the ground and the downward gravitational force (\( F_{G} = mg \)). This diagram helps to clearly show the direction of forces and how they interact.

In the diagram for the froghopper, since it leaps upwards, the normal force arrow points upwards while the weight arrow points downwards. Both these forces play a crucial role in understanding how the insect manages to achieve its jump, as the net force arises from the difference between these two forces.

Force Calculation

Calculating forces involves using Newton’s second law and understanding kinematics. In this task, the force exerted by the ground on the froghopper during a jump is calculated by determining the net force needed for its upward acceleration.

• First, calculate the net force using the formula \( F_{net} = ma \).
• Then add the gravitational force to the net force to find the normal force.
• Express the normal force relative to the frog's weight.

These steps enable us to find that the ground exerts a force of approximately 49.319 N. This force is significantly larger, around 408.75 times, than the froghopper’s own weight. This demonstrates the powerful interaction of the forces at play in such an extraordinary jump.