WORK, ENERGY AND POWER

CHAPTER 5 WORK, ENERGY AND POWER 5.1

Work done by constant force

5.2

Kinetic Energy and Work Energy Principle

5.3

Gravitational Potential Energy

5.4

Mechanical Energy and Its Conservation

5.5

Power

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5.1

Work done by constant force

WORK Definition: Work done on an object by a constant force is the product of the component of force, acting parallel to the displacement of the object times the magnitude of the displacement.

W  Fs W  F cos  s

SI unit : Joule (J) 1 J = 1 Nm

F θ

F cos θ

Position 1 ( Initial )

Position 2 ( Final )

s (displacement)

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Cases of work done by constant force Case 1 : – work done by a horizontal force, F on an object.

 F

θ 0



 s

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W  Fs cos θ W  Fs

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Case 2 : – work done by a horizontal forces, F1 and F2 on an object.

 F1  F2

W1  F1 s cos 0

 s



 W W

1

W2  F2 s cos 0

 W2  F1 s  F2 s 

 W F  F s 1

2

 W F s net

SF017

4



Case 3 : – work done by a vertical force, F on an object.

 F

θ  90



 s

SF017

W  Fs cos θ W 0 J

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Case 4 : – work done by a force, F and frictional force, f on an object.

 F

 f



Wnet  Fnet s

 s Fnet  F cos   f

Wnet  F cos   f s

ma

SF017

Wnet  mas

6

•

Caution :

– Work done on an object is zero when

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F = 0 or s = 0 and  = 90.

Example 1 : Find the work done if the force is 45.0 N, the angle is 50° and the displacement is 75.0 m.

Solution :

Answer : 2169.41 J

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Example 2 :

You push your physics reference book 1.50 m along a horizontal table with a horizontal force of 5.00 N. The frictional force is 1.60 N. Calculate a. the work done by the 5.00 N force, b. the work done by the frictional force, c. the total work done on the book.

f  1.60 N

F  5.00 N s  1.50 m

Solution :

Answer : 7.5 J, -2.4 J, 5.1 J

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Example 3 : A box of mass 20 kg moves up a rough plane which is inclined to the horizontal at 25.0ď‚°. It is pulled by a horizontal force F of magnitude 250 N. The coefficient of kinetic friction between the box and the plane is 0.300. a. If the box travels 3.80 m along the plane, determine i. the work done on the box by the force F, ii. the work done on the box by the gravitational force, iii. the work done on the box by the reaction force, iv. the work done on the box by the frictional force, v. the total work done on the box. b. If the speed of the box is zero at the bottom of the plane, calculate its speed when it is travelled 3.80 m.

(Given g = 9.8 m s2)

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Solution :

Answer : 950 J, 0 J, 0 J, -202.51 J, 747.49 J

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Work: Area under the F-versus-x Curve

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suhaida dila 2012

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Example 4 :

F (N)

5 0

4

3

5

6

7

s (m)

A horizontal force F is applied to a 2.0 kg radio-controlled car as it moves along a straight track. The force varies with the displacement of the car as shown in figure above. Calculate the work done by the force F when the car moves from 0 to 7 m.

Solution :

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Answer : 18 J

ENERGY  is defined as the system’s ability to do work.  S.I. unit : joule, J  scalar quantity.

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5.2

Kinetic Energy and Work Energy Principle DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass m and speed v is given by

KE  mv



1 2

2



W  mas   m v  v  mv  mv 1 2

v 2f  vo2  2ax 

2 f

2 o

1 2

2 f

1 2

2 o

ax  12 v 2f  vo2  16

THE WORK-ENERGY THEOREM Work done by an object equal to change in kinetic energy.

When a net external force does work on and object, the kinetic energy of the object changes according to

W  KEf  KEi  mv  mu 1 2

2

1 2

2

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Example 5 : The mass of the space probe is 474 kg and its initial velocity is 275 m/s. If the 56.0 mN force acts on the probe through a displacement of 2.42Ă—109 m, what is its final speed?

Solution :

Answer : 805 m/s

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5.3

Gravitational Potential Energy

DEFINITION OF POTENTIAL ENERGY The energy stored in a system which depends on the relative position of the body in the system.

Wgravity  Fs  mgho  h f 

PE  mgh THE PRINCIPLE OF CONSERVATION OF ENERGY Energy can neither be created not destroyed, but can only be converted from one form to another.

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Example 6 :

A 65 kg firefighter climbs a flight of stairs 20 m high. How much work is required?

Solution :

Answer : 12740 J

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Elastic Potential Energy HOOKE’S LAW

F  kx where k = spring constant ∆x = extension

1 2 EPE  kx 2

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Example 7 : A spring has a spring stiffness constant of 440 N/m. How much must this spring be stretched to store 25 J of potential energy?

Answer : 0.34 m

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5.4

Mechanical Energy and Its Conservation

The total mechanical energy (E = KE + PE) of an object remains constant as the object moves, provided that the net work done by external nononservative forces is zero.

( PE  KE )i  ( PE  KE ) f

WORK-ENERGY THEOREM

If a surface is rough (friction force is exist), there are ENERGY LOST.

So WORK-ENERGY THEOREM become

( PE  KE )i  ( PE  KE ) f  E.L

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The Conservation of Mechanical Energy

So WORK-ENERGY THEOREM become

( PE  KE )i  ( PE  KE ) f

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Example 8 : The gymnast leaves the trampoline at an initial height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?

Answer : 8.4 m/s

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Example 9 : A stationary object is at point A at 3 m above ground initially. It is set to move and arrives at B, the ground level with speed 4 m/s.

Determine the percentage of energy loss due to friction.

Solution :

Answer : 73%

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THE PRINCIPLE OF CONSERVATION OF ENERGY

Energy can neither be created not destroyed, but can only be converted from one form to another.

Pendulum Bob Ep (max)

Ep (max)

Ep + Ek

Ep + Ek Ek (max)

Equilibrium Position 27

Example 10 :

A pendulum bob with mass 200 g attaching to a light string was displaced to height of 20 cm from the equilibrium position. When the bob is released, the swinging of pendulum starts.

Calculate: a) Maximum velocity b) Velocity at height 5 cm from equilibrium position c) Kinetic energy at height 5 cm from equilibrium Solution :

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Solution :

Answer : 2 m/s, 1.73 m/s, 0.299 J

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Example 11 :

C

v0

A 10 m

B

D

A ball of mass 0.50 kg is at point A with initial speed, v0=4 m s-1 and at a height of 10 m as shown in figure above .(Ignore frictional force). Calculate a. the total energy at point A, b. the speed of the ball at point B where the height is 3 m, c. the speed of the ball at point D, d. the maximum height of point C so that the ball can pass over it.

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Solution :

Answer : 53 J, 12.38 m/s, 14.56 m/s, 10.82 m

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Example 12 : A block of mass 0.2 kg is tied to one end of a spring of constant 5.0 N/m.

The other end is hinged to a wall. The block is pulled and the spring stretched 0.1 m. The system then is released. What is the speed of the block at x = 0.08 m? Friction is neglected.

Solution :

Answer : 0.3 m/s

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5.5

Power

POWER DEFINITION OF POWER Average power is the rate at which work is done, and it is obtained by dividing the work by the time required to perform the work.

Work W Fs P    Fv Time t t

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Example 13 : A motor of power 5 kW can lift an object vertically upward with constant speed through 35 m in 10 s. Determine the mass of object.

Solution :

Answer : 146 kg

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Example 14 :

The power of the engine of a car of mass 950 kg is 3.5 kW when the car moves at a constant speed of 20 m/s along a straight road. What is the total resistance (force) against the car at this speed?

Solution :

Answer : 175 N

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