Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

Answers without the blur.

Just sign up for free and you're in.

Short Answer

Calculate the power output in watts and horsepower of a shot-putter who takes \(1.20{\rm{ s}}\) to accelerate the \(7.27 - {\rm{kg}}\) shot from rest to \(14.0{\rm{ m}}/{\rm{s}}\), while raising it \(0.800{\rm{ m}}\). (Do not include the power produced to accelerate his body.)

The average power output of the shot-putter is \(641.26{\rm{ W}}\) or \(0.86{\rm{ hp}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of Concepts

The net work done on the system is given as,

\(W = \Delta KE + \Delta PE\)

Here, \(\Delta KE\) is the change in kinetic energy, and \(\Delta PE\) is the change in potential energy.

The power output of the system is given as,

\(P = \frac{W}{T}\)

Here, \(T\) is the time required to do the work.

Step 2: Calculate the power output of shot-putter in watts and horsepower

The change in kinetic energy is given as,

\(\Delta KE = \frac{1}{2}m\left( {v_f^2 - v_i^2} \right)\)

Here, \(m\) is the mass of the shot \(\left( {m = 7.27{\rm{ kg}}} \right)\), \({v_f}\) is the final velocity of the shot \(\left( {{v_f} = 14.0{\rm{ m}}/{\rm{s}}} \right)\), and \({v_i}\) is the initial velocity of the shot (\({v_i} = 0\) as the shot starts from rest).

Putting all known values,

\(\begin{array}{c}\Delta KE = \frac{1}{2} \times \left( {7.27{\rm{ kg}}} \right) \times \left[ {{{\left( {14.0{\rm{ m}}/{\rm{s}}} \right)}^2} - {{\left( 0 \right)}^2}} \right]\\ = 712.46{\rm{ J}}\end{array}\)

The change in potential energy is given as,

\(\Delta PE = mg\left( {{h_f} - {h_i}} \right)\)

Here, \(m\) is the mass of the shot \(\left( {m = 7.27{\rm{ kg}}} \right)\), \(g\) is the acceleration due to gravity \(\left( {g = 9.81{\rm{ m}}/{{\rm{s}}^2}} \right)\), \({h_f}\) is the height of the shot lifted \(\left( {{h_f} = 0.800{\rm{ m}}} \right)\), and \({h_i}\) is the initial height of the shot \(\left( {{h_i} = 0} \right)\).

Putting all known values,

\(\begin{array}{c}\Delta PE = \left( {7.27{\rm{ kg}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left[ {\left( {0.800{\rm{ m}}} \right) - \left( 0 \right)} \right]\\ = 57.05{\rm{ J}}\end{array}\)

The total work done by the shot-putter can be calculated using equation (1.1).

Putting all known values into equation (1.1),

\(\begin{array}{c}W = \left( {712.46{\rm{ J}}} \right) + \left( {57.05{\rm{ J}}} \right)\\ = 769.51{\rm{ J}}\end{array}\)

The shot-putter takes a time of \(t = 1.20{\rm{ s}}\) to do the work.

The power output of the shot-putter can be calculated using equation (1.2).

Putting all known values into equation (1.2).

\(\begin{array}{c}P = \frac{{769.51{\rm{ J}}}}{{1.20{\rm{ s}}}}\\ = 641.26{\rm{ W}}\end{array}\)

Converting power output of the shot-putter from watts to horsepower.

\(\begin{array}{c}P = \left( {641.26{\rm{ W}}} \right) \times \left( {\frac{{1{\rm{ hp}}}}{{746{\rm{ W}}}}} \right)\\ = 0.86{\rm{ hp}}\end{array}\)

Therefore, the required average power output of the shot-putter is \(641.26{\rm{ W}}\) or \(0.86{\rm{ hp}}\).

Find Study Materials for

Create Study Materials

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Calculate the power output in watts and horsepower of a shot-putter who takes \(1.20{\rm{ s}}\) to accelerate the \(7.27 - {\rm{kg}}\) shot from rest to \(14.0{\rm{ m}}/{\rm{s}}\), while raising it \(0.800{\rm{ m}}\). (Do not include the power produced to accelerate his body.)

Step by Step Solution

Step 1: Definition of Concepts

Step 2: Calculate the power output of shot-putter in watts and horsepower

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Calculate the power output in watts and horsepower of a shot-putter who takes \(1.20{\rm{ s}}\) to accelerate the \(7.27 - {\rm{kg}}\) shot from rest to \(14.0{\rm{ m}}/{\rm{s}}\), while raising it \(0.800{\rm{ m}}\). (Do not include the power produced to accelerate his body.)

Step by Step Solution

Step 1: Definition of Concepts

Step 2: Calculate the power output of shot-putter in watts and horsepower

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics