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Estimate the average force exerted by a shot-putter on a \(7.0-\mathrm{kg}\) shot if the shot is moved through a distance of \(2.8 \mathrm{~m}\) and is released with a speed of \(13 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The average force exerted is 211.25 N.

Step by step solution

01

Understand the Problem

We need to find the average force exerted by the shot-putter. We can use the kinematic equations and Newton's second law. The work-energy principle will also be useful here, stating that work done is equal to the change in kinetic energy.
02

Identify Given Values

We have the mass of the shot, which is \( m = 7.0 \) kg, the distance it is moved \( d = 2.8 \) m, and the final speed \( v = 13 \) m/s. We assume the initial speed is 0 since it's not stated otherwise.
03

Calculate Work Done

The work done on the shot is equal to the change in kinetic energy. The final kinetic energy \( KE = \frac{1}{2}mv^2 \). Substitute \( m = 7.0 \) kg and \( v = 13 \) m/s: \( KE = \frac{1}{2} \times 7.0 \times (13)^2 \).
04

Evaluate Change in Kinetic Energy

Calculate the kinetic energy: \( KE = 0.5 \times 7.0 \times 169 = 591.5 \) Joules. This is the total work done by the force.
05

Apply Work-Energy Principle

By the work-energy principle, work done \( W = F \cdot d \), where \( F \) is the average force and \( d \) is the distance moved. So, \( 591.5 = F \times 2.8 \).
06

Solve for Average Force

Isolate the force \( F \): \( F = \frac{591.5}{2.8} \).
07

Calculate the Average Force

Perform the division: \( F = 211.25 \) N.

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Key Concepts

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

Average Force
The concept of average force is crucial in understanding how forces act over a displacement to produce motion. The average force is essentially the total force exerted over a particular distance divided by that distance. In simple terms, it represents the "standard" amount of force being consistently used to move an object over the given distance.

In the context of our exercise, the shot-putter applies force to move a shot through a specific distance before releasing it. To find the average force, we utilize the work-energy principle, which tells us that the work done by the force is equal to the change in kinetic energy of the object. While calculating, we already know the work done (which is the change in kinetic energy) and the distance. Hence, we only need to divide the work by the distance to find the average force:
  • Given: Work done = 591.5 Joules
  • Distance = 2.8 meters
This results in an average force achieved as 211.25 N. This calculation is not only theoretical but reflects the practical force throughout the distance, smoothing out any variations the shot-putter might actually apply.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. The faster the object moves, the more kinetic energy it has. It is calculated using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.

In the exercise, the shot was initially at rest and later reached a velocity of 13 m/s when released. Hence, the kinetic energy at the point of release becomes important for further calculations. Because the initial kinetic energy was zero (due to zero velocity), the change in kinetic energy is the same as the final kinetic energy we calculate:
  • Mass \( m = 7.0 \text{ kg} \)
  • Velocity \( v = 13 \text{ m/s} \)
The final kinetic energy is determined using: \[ KE = \frac{1}{2} \times 7.0 \times 13^2 = 591.5 \text{ Joules} \]

This calculation tells us exactly how much energy was converted into motion from the forces applied, which links directly to the work-energy principle used alongside average force.
Newton's Second Law
Newton's Second Law fundamentally describes how the velocity of an object changes when it is subject to an external force. It is generally formulated as \( F = ma \), where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration.

However, in solving this exercise, we explored an additional application of the law related to work and energy, rather than direct application of \( F = ma \). The force exerted by the shot-putter resulted in acceleration that changed the shot's speed. Although the law is not explicitly calculated through acceleration here, it underpins the fundamental understanding that force impacts the change in velocity (and thus kinetic energy) of the shot.

The work-energy principle acts as an extended arm of Newton's laws, particularly useful for cases where the force might not be constant or easily measured. Instead, we focus on the result of the exerted force (i.e., the work done and the energy change), giving us a practical understanding and application of the second law in terms of energy transformations over time.

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Most popular questions from this chapter

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