91Ó°ÊÓ

In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about \(42^{\circ}\) ) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a \(7.260 \mathrm{~kg}\) shot is accelerated along a straight path of length \(1.650 \mathrm{~m}\) by a constant applied force of magnitude \(380.0 \mathrm{~N}\), starting with an initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) \(30.00^{\circ}\) and (b) \(42.00^{\circ}\) ? (Hint: Treat the motion as though it were along a ramp at the given angle.) (c) By what percentage is the launch speed decreased if the athlete increases the angle from \(30.00^{\circ}\) to \(42.00^{\circ}\) ?

Step by step solution

01

Calculate work done by the force

First, calculate the work done by the applied force on the shot. The formula for work done is given by:\[ W = F \times d \times \cos(\theta) \]where \( F = 380.0 \text{ N} \) is the magnitude of the force, \( d = 1.650 \text{ m} \) is the length of the path along which the force acts, and \( \theta \) is the angle between the direction of the force and the direction of motion. For the given scenario, the applied force is directly along the path, therefore \( \theta = 0 \) and \( \cos(0) = 1 \). Hence,\[ W = 380.0 \times 1.650 \times 1 = 627.0 \text{ J} \]

02

Apply work-energy principle

Use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy:\[ \Delta KE = W = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i \]where \( m = 7.260 \text{ kg} \) is the mass, \( v_i = 2.500 \text{ m/s} \) is the initial speed, and \( v_f \) is the final speed. Rearrange to find \( v_f \):\[ 627.0 = \frac{1}{2} \times 7.260 \times v_f^2 - \frac{1}{2} \times 7.260 \times (2.500)^2 \]

03

Solve for final speed at angle 30°

Substitute \( v_i = 2.500 \text{ m/s} \) and solve for \( v_f \) when the angle is \( 30.00^{\circ} \). The elevation component of motion does not change the calculation because energy is conserved regardless of angle on a ramp. Simplify and solve:\[ 627.0 = 3.63 \times v_f^2 - 22.625 \]\[ 649.625 = 3.63 \times v_f^2 \]\[ v_f^2 = \frac{649.625}{3.63} \]\[ v_f \approx 13.41 \text{ m/s} \]

04

Solve for final speed at angle 42°

Use the same approach as in Step 3, since energy conservation applies similarly:\[ 627.0 = 3.63 \times v'_f{}^2 - 22.625 \]\[ 649.625 = 3.63 \times v'_f{}^2 \]\[ v'_f{}^2 = \frac{649.625}{3.63} \]\[ v'_f \approx 13.41 \text{ m/s} \]Note: Since the only difference in physics is the effect of angle on practical performance, not the calculated result, \( v_f \) is the same in this idealized analysis for both angles.

05

Calculate percentage decrease in launch speed

Since both ideal ramps yield the same final speed in theoretical work-energy conservation:\[ \text{Percentage decrease} = \left(\frac{v_f'(\theta = 42^{\circ}) - v_f(\theta = 30^{\circ})}{v_f(\theta = 30^{\circ})}\right) \times 100\% \]\[ \text{Percentage decrease} = \left(\frac{13.41 - 13.41}{13.41}\right) \times 100\% = 0\% \]The practical angle variation does not impact conserved ideal ramp launch speed, leading to no theoretical decrease.

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Key Concepts

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

Work-Energy Principle

The work-energy principle connects how work done on an object relates to its kinetic energy. When a force acts over a distance, it does work, which can change the kinetic energy of that object. This principle is important for understanding projectile motion in physics, like in a shot put throw.

• Work Done: Work is calculated as the product of the force applied, the distance over which it is applied, and the cosine of the angle between the force and direction of motion. In formal terms, it is given by the equation \( W = F \times d \times \cos(\theta) \).
• Kinetic Energy: It’s the energy an object possesses due to its motion. The work-energy principle states the work done on the object equals the change in its kinetic energy, expressed as \( \Delta KE = \frac{1}{2} m v^2_f - \frac{1}{2} m v^2_i \), where \( m \) is the mass, \( v_i \) is the initial velocity, and \( v_f \) is the final velocity.
• Practical Application: In shot putting, this principle helps calculate how much energy is transferred to the shot through the athlete's effort, allowing us to determine its speed along a certain path.

This principle helps bridge the gap between theoretical motion and real-life applications in sports and other areas of physics.

Acceleration

Acceleration measures how quickly an object's velocity changes. In the context of projectile motion, understanding acceleration is crucial to determine the ultimate speed of an object like a shot being thrown.

• Definition: Acceleration is the rate at which velocity changes with time. If a shot putter exerts a consistent force, the shot undergoes a steady acceleration.
• Calculation: Given a constant applied force, the acceleration \( a \) can be determined using Newton's second law \( F = ma \), where \( F \) is the force applied and \( m \) is the mass of the object.
• Impact in Shot Put: During the acceleration phase, the shot's speed increases until the force is no longer applied. This increase in speed determines the shot's travel distance.

In essence, knowing how acceleration affects motion allows us to predict the shot's speed and trajectory when thrown.

Launch Angle

The launch angle is pivotal in projectile motion and deeply affects how far an object can travel. It's why shot putters often have different angles than theoretical models suggest.

• Theoretical Optimum: Physics suggests an angle of about 42° provides the maximum horizontal distance if speed and height are constant.
• Practical Considerations: Athletes might choose different angles due to factors like muscle efficiency or technique. A smaller launch angle can sometimes yield better results by enhancing the force application and speed.
• Example: The shot-put exercise explores how adjusting from a 30° to 42° angle affects launch speed, highlighting angle's importance in optimizing performance.

Understanding launch angles is vital for athletes and engineers alike, affecting everything from sports performance to missile trajectory.

Physics Problems

Physics problems can challenge your understanding, requiring application of various concepts. The shot put problem is an example that ties multiple physics concepts together.

• Concepts Involved: Problems often involve applying work-energy principles, calculating acceleration, and considering launch angles—all crucial to solving projectile motion problems like the shot put.
• Problem-Solving Steps: Start by identifying known values and equations, calculate relevant quantities like work done or speed, and apply principles like energy conservation to find unknowns.
• Real-world Connections: Solving such problems helps understand real-world motions, from athletic events to engineering challenges.

By breaking down physics problems into steps and understanding underlying concepts, you can solve even complex issues efficiently and accurately.