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How much work is required for a \(73-\mathrm{kg}\) runner to accelerate from rest to a speed of \(7.5 \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
The work required is 2053.125 Joules.

Step by step solution

01

Identify the Formula for Work

To find out how much work is needed, we use the formula for kinetic energy: \[ W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \] where \(m\) is the mass, \(v\) is the final velocity, and \(u\) is the initial velocity.
02

Substitute Known Values

We know the mass \(m = 73\,\text{kg}\), the initial velocity \(u = 0\,\text{m/s}\), and the final velocity \(v = 7.5\,\text{m/s}\). Substitute these into the kinetic energy formula: \[ W = \frac{1}{2} \times 73 \times (7.5)^2 - \frac{1}{2} \times 73 \times (0)^2 \]
03

Calculate the Final Kinetic Energy

Compute the final kinetic energy term:\[ \frac{1}{2} \times 73 \times (7.5)^2 = \frac{1}{2} \times 73 \times 56.25 = 2053.125\,\text{J} \]
04

Calculate the Initial Kinetic Energy

Since the initial velocity is zero, the initial kinetic energy is zero:\[ \frac{1}{2} \times 73 \times (0)^2 = 0\,\text{J} \]
05

Calculate the Work Done

Find the work done to accelerate the runner from rest to 7.5 m/s by finding the difference in kinetic energy:\[ W = 2053.125 - 0 = 2053.125\,\text{J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is one of the most fundamental concepts in physics. It describes the energy that an object possesses due to its motion. To calculate kinetic energy (KE), you can use the formula:
  • \[ KE = \frac{1}{2} m v^2 \]
Here, \(m\) represents the mass of the object and \(v\) stands for its velocity. The velocity is squared, meaning that even small increases in velocity can result in large increases in kinetic energy.

Consider a runner accelerating from a stationary position: initially, they have zero kinetic energy because they aren't moving. As they pick up speed, they gain kinetic energy. It's important to understand that this gain is not just because they're moving, but because the velocity's influence on kinetic energy is exponential due to the square factor in the formula.

Remember, kinetic energy is always a positive quantity because mass and the square of velocity are non-negative numbers. This energy is what allows the runner to keep moving once enough energy (or work) has been applied to overcome inertia.
Mass and Velocity
In physics, mass and velocity are crucial elements that directly impact an object's kinetic energy. The mass is a measure of how much matter the object contains, and it directly scales the energy. Heavier objects require more energy to move at the same speed as lighter objects.

Velocity is equally essential, but its effect on kinetic energy is even more pronounced due to the formula's square term. If you double an object's velocity, its kinetic energy increases by a factor of four. This relationship highlights how changes in speed dramatically affect the energy involved.

When a runner with a mass of 73 kg accelerates from rest to 7.5 m/s, the change in speed is the key player in calculating how much work is required. The mass stays constant, but as velocity increases, the energy needed climbs significantly. Understanding the roles of mass and velocity helps us grasp why athletes of different sizes and speeds experience varied energy demands.
Work-Energy Principle
The work-energy principle is a fundamental concept that connects work and energy. It expresses that the work done on an object results in a change in its kinetic energy. The principle can be summarized in the formula:
  • \[ W = \Delta KE \]
Here, \(W\) represents work, and \(\Delta KE\) is the change in kinetic energy. If an object starts from rest, the initial kinetic energy is zero, simplifying the calculations.This principle is essential for problems where forces are doing work over a distance, like when a runner pushes themselves forward on a track. The work done by their muscles results in a change from no motion to a velocity of 7.5 m/s.

In our example, the total work needed to bring the runner from rest to the specified speed is calculated based solely on the change in kinetic energy, demonstrating why understanding this principle is crucial. By gaining a deeper comprehension of how work transforms into kinetic energy, you can solve a range of practical physics problems.

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