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Potential Energy and Work

 

Table of Contents
Introduction
Problems
Answers

  

Introduction

Potential Energy is stored energy.  It can also be defined as energy that is due to forces dependent on position or configuration of a body.

Potential energy comes in many forms.  Each has its own definition of potential energy.

gravitational        
potential
energy		 If an object if lifted from the floor to the top of a table, the work done on the object is equal to the gain in potential energy.

W grav = Fd = mgh                      h is the height above a reference level. 

A reference level must be picked when working problems.  It must be picked before you start the problem and cannot be changed in the middle of the problem.


In the diagram to the left, the mass will first be raised from the ground at h1to h2.  It is then dropped from h2 and reaches the ground at h1.  When it reaches the ground it drives the stake into the ground.

When raised from the ground

W grav = PE2  -  PE1     

W grav = mgh2  -  mgh1 

Work must be done on the object to pick it up and move it from position 1 to position 2.

     
When it is dropped, the object converts PE into KE.  The potential energy is constantly decreasing and the potential energy is constantly increasing.  

KE = 1/2 mv2 = - /\PE    assuming it starts from rest 

   
It uses the KE to pound the stake into the ground.

   

    

     
elastic
potential
energy		 Hooke's Law shows a linear relationship between the applied force and the amount a spring is stretched or compressed. 

The area under the line is a right triangle.  The equation for the area is

A = W = PE elastic = 1/2 kx2 

 
Table of Contents

     

Problems

1.      Compared to the ground floor, what is the potential energy of a basketball, mass of 2.50 kg, that is lifted to the top of the Empire State Building, height is 315 m?  If the ball is dropped, how fast is it traveling when it hits the ground?
   
2. How high will a 400. g ball be raised from ground level if 5.00 x 10 3 J of work are used to lift the ball?  If the ball is dropped, what will its kinetic energy be when it hits the ground?
   
3. 900. N of force are necessary to compress a spring 35.0 cm.  How much work was necessary to compress the spring.  If the spring is used to propel a 200. g ball, how fast does the ball travel when it leaves the spring?
   
4. 600. N of force are necessary to compress a spring 15.0 cm.  If the spring is used to propel a 300. g ball at 15.0 m/s, how much must the spring be compressed?

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Answers

1.      Compared to the ground floor, what is the potential energy of a basketball, mass of 2.50 kg, that is lifted to the top of the Empire State Building, height is 315 m?  If the ball is dropped, how fast is it traveling when it hits the ground?
   
   
PE = mgh = 2.50 kg x 9.80 m s -2 x 315 m = 7720 J

KE = 1/2 mv2                Potential Energy at the top is completely converted into KE at the bottom.
   

         2 KE                 2 x 7720 J
v = ( ---------) 1/2 = ( ------------------) 1/2 = 78.6 m/s
            m                      2.50 kg
    
   

   
2. How high will a 400. g ball be raised from ground level if 5.00 x 10 3 J of work are used to lift the ball?  If the ball is dropped, what will its kinetic energy be when it hits the ground?
   
   
                                W              5.00 x 10 3 J
W = mgh  =>  h = ------- = ------------------------------ = 1280 m
                               mg        0.400 kg x 9.80 m s -2    
   

KE when the ball hits the ground is equal to the PE at the top which is equal to the Work necessary to lift the ball.   KE = 5.00 x 10 3 J
  
   

   
3. 900. N of force are necessary to compress a spring 35.0 cm.  How much work was necessary to compress the spring.  If the spring is used to propel a 200. g ball, how fast does the ball travel when it leaves the spring?
   
   
                           F       900. N
F = kx =>  k = ----- = ------------- = 2570 N/m                    First calculate value of k.
                           x        0.350 m
   

W = 1/2 kx2 = 1/2 x 2570 N/m x (0.350 m)2 = 157 J       This is equal to the KE when the ball is released.

KE = 1/2 mv2                                                                     Solve for the velocity.
   

         2 KE                 2 x 157 J
v = ( ---------) 1/2 = ( ------------------) 1/2 = 39.6 m/s
            m                      0.200 kg
    
   

   
4. 600. N of force are necessary to compress a spring 15.0 cm.  If the spring is used to propel a 300. g ball at 15.0 m/s, how much must the spring be compressed?
   
    
                           F       600. N
F = kx =>  k = ----- = ------------- = 4.00 x 10 3 N/m         First calculate value of k.
                           x        0.150 m
   

KE of ball when released is equal to the work necessary to compress the spring.
   

                                                 mv2                 0.300 kg x (15.0 m s -1)2  
1/2 mv2 = 1/2 kx2   =>  x = (--------- ) 1/2  = ( --------------------------------- ) 1/2  =  0.130 m = 13.0 cm
                                                   k                         4.00 x 10 3 N m -1 
   

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