Problem 9 - College Physics

Question

Animal dynamics. An adult 68 \(\mathrm{kg}\) cheetah can accelerate from rest to 20.1 \(\mathrm{m} / \mathrm{s}(45 \mathrm{mph})\) in 2.0 \(\mathrm{s}\) . Assuming constant acceleration, (a) find the net external force causing this acceleration. (b) Where does the force come from? That is, what exerts the force on the cheetah?

Step-by-Step Solution

Answer

(a) The net external force is 683.4 N. (b) The force comes from the ground, as the cheetah pushes against it.

Step 1: Identify the Known Values

The problem provides the following information: the mass of the cheetah, \(m = 68 \, \mathrm{kg}\), the final velocity, \(v_f = 20.1 \, \mathrm{m/s}\), the initial velocity, \(v_i = 0 \, \mathrm{m/s}\) (because the cheetah starts from rest), and the time it takes to reach this velocity, \(t = 2.0 \, \mathrm{s}\).

Step 2: Calculate the Acceleration

Since acceleration \(a\) is constant, we use the formula \(a = \frac{v_f - v_i}{t}\). Substituting the known values, we find:\[a = \frac{20.1 \, \mathrm{m/s} - 0 \, \mathrm{m/s}}{2.0 \, \mathrm{s}} = 10.05 \, \mathrm{m/s^2}\]

Step 3: Apply Newton's Second Law

Now that we have the acceleration, apply Newton's Second Law to find the force. According to the law, \(F = m \times a\). Using the known values:\[F = 68 \, \mathrm{kg} \times 10.05 \, \mathrm{m/s^2} = 683.4 \, \mathrm{N}\]

Step 4: Determine the Source of the Force

The net external force on the cheetah comes from the ground. When the cheetah pushes back against the ground with its legs, by the third law of motion (action and reaction), the ground exerts an equal and opposite force on the cheetah, propelling it forward.

Key Concepts

Constant Acceleration

When we talk about constant acceleration, we're referring to a situation where the rate of change of velocity of an object remains steady over time. In other words, the object speeds up or slows down at the same rate throughout the period of interest. This is often represented by the formula:

• \(a = \frac{v_f - v_i}{t}\)

Here, \(a\) is the constant acceleration, \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(t\) is the time taken.
For the cheetah in our scenario, it starts from rest (meaning its initial velocity \(v_i\) is 0), and speeds up to 20.1 meters per second in just 2 seconds. During this time, its acceleration does not change, and we calculated it to be \(10.05 \, \mathrm{m/s^2}\).
This concept is quite useful in physics, as it simplifies the math – knowing just two velocities and the time, we can easily find how quickly the object's speed is changing.

Force Calculation

Force calculation is a practical application of Newton's Second Law of Motion. This law states that the force acting upon an object is equal to the mass of the object times its acceleration, described mathematically by:

• \(F = m \times a\)

In our example, the cheetah has a mass of 68 kg and, from our previous calculation, an acceleration of \(10.05 \, \mathrm{m/s^2}\).
By multiplying mass and acceleration, we find that the external force needed to accelerate the cheetah is \(683.4 \, \mathrm{N}\) (Newtons).
Understanding force calculations helps in determining how much effort is required to change the motion of an object. It's useful in everyday applications, like understanding how engines move cars or how athletes increase speed when running.

Action and Reaction

Action and reaction are fundamental concepts from Newton's Third Law of Motion, which states that every action has an equal and opposite reaction.
This means that forces always come in pairs. Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force back on the first object.

• Consider the cheetah: As it pushes against the ground with its legs (action), the ground pushes back with an equal force in the opposite direction (reaction).

This reaction from the ground is what actually propels the cheetah forward.
This principle is everywhere in our daily lives – from walking (our feet pushing against the ground) to swimming (pushing water backwards so that we move forward).
Understanding action and reaction helps explain why objects move or stay at rest and is crucial in fields like engineering and biomechanics.