FE Physics
 Formulas & Explanation
 
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Work and energy

 

Work

                                                                                                

W = F s cos α                    

 

W   work in J (Joule)  (not to be confused 

       with the weight)

F     force in N

s     displacement in m

α    angle between  F en s

 

 

 

Example 1

See the figure. On an object acts a force F. The object moves 100 m to the right.

The angle α between F and the x-direction is 35 o. The weight is 30 N.

Also work on this object a frictional force (f=5.0 N) and a normal force Fn.

 

a.Find the work of F

Resolve  F in the x-direction (parallel to the displacement) and in the y-direction (perpendicular to the displacement)

 

F x = 16.4 N         (Calculation : cos 35 o = Fx/ F         Fx = F cos 35 o)       

WF = Fx s = (16.4)( 100) = 1640 J

 

Also possible :  W = F s cos α = (20)( 100 cos 35) = 1640 J

 

Remark        

If force and displacement have the same direction :  W > 0 J

 

b. Find the work of the frictional force f

Wf = - f s= - (5.0)(100) = - 500 J

of

Wf = f s cos α =(5.0)( 100 cos 180o) = (5.0)( 100)( -1) = - 500 J

( If force and displacement are oppositely directed then  W < 0 J)

 

c. Find the work of weight and normal force

W w= W Fn = 0 J

( If F s then : W = 0 J,       α= 90o    cos 90o = 0)

 

Example 2          

 An object with a mass of 5.0 kg falls  20.0 m downward.

How much is the work of the weight ?

 W = m g = (5.0)( 9.81) = 49.05 N

 Ww= W s = (49.05)( 20.0) = 981 J = 9.8 x 102 J

 

Example 3

 we pull a spring ( k = 50 Nm -1) 20 cm.

Calculate the work done by the force excert by the spring.

Method 1

If the spring has been pulled to the right, the force in the spring is directed to the left.

Thus the work of Fspring is  negative .

WFspring = - ½ k x2  = - ½ (50)(0.20)2 = - 1.0 J

                                                                                                             

Method 2

Determine work with  F versus x graph        

F = k x = (50)( 0.20) = 10.0 N

 

 

 

 

 

 

W  is equal to the arae under the graph of F versus x

W  = ½ (10.0)( 0.20) =  - 1,0 J.

W < 0 because the force of the spring and displacement are oppositely directed.

 

 

Mechanica  energy

Kinetic energy                                                                   :  Ek    = ½ m v2                    

Gavitational potential energy                                     :  Eg    = m g h      

Potential energy in a spring                                         :  Es = ½ k x 2    (Energy stored in a pulled or 

                                                                                                                          pressed spring)

 

Conservation of Mechanical Energy 

                              

Falling body without air resistance

 

Example 4

From a height of 15 m an object with mass of 2.0 kg  is thrown down with an initial velocity of  3.0 ms-1 .

Find the velocity at which the object the ground reaches.

 

The mechanical energy is constant, because there is no friction.

 

Ek + Eg = constant.  

The heighst point is called H and lowest point G

(Ek + Eg) H  =( Ek + Eg) G

  ½ (2.0) (3.0)2 + (2.0)( 9.81)( 15) = Ek + 0        (h = 0 m)

9.0 + 294.3 = Ek,G  

E k,G = 303.3 J

E k,G = ½  (2.0) (v)2

303.3 = ½ (2.0)(v)2

v = 17.4 m/s

     

Remark

Because the mass is a factor in every term , the mass can be eliminated.

We can write:

   ½ (vH)2 + g hH = ½ (vG)2

 

 

The  Work – Energy Theorem

 

∑ W = Δ Ek

 

Example  5   

 

               

          

                 

On an object act a force F of60 N and a frictional force of 20 N

The object goes to the right. The initial velocity is 4.0 m s-1

Calculate the velocity v2   at a displacement of 30 m (see drawing)

 

∑ W = E k,2  - E k,1

Fnet  s = ½ m v 22 – ½ m v12

(40 )( 30) = ½ (2.0)( v2)2 – ½ (2.0)( 4.0)2

1200 = v22 – 16

v22 = 1216

v2 = 34.9 m/s

 

Power and work

Example 6

 

A car with mass 1250 kg drives with a constant speed of 90 km h-1

The petrol consumption of the car is 6.7 liter petrol per 100 km.

The heat of combustion of the petrol is 33 x 109 J m-3

The frictional force is 500 N

Find the efficiency of the motor in the car.

 

Echem = (6.7)( 33 x 106 )= 2.21 x 108  J

Velocity is constant. Therefore  : Fm = f = 500 N

 

WFm = Fm  s =( 500)(100 x 103) = 5.0 x 107 J

 

Efficiency η = (W/E)  100 %= (5.0 x 107)/(2.21.108)  100 % = 0.23 x 100 % = 23 %

        

Example 7

Calculate the power  P  of the car at this velocity of 90 kmh-1

Solution1 : The time required for 100 km is  100/90 houres  = 1.11 h = 4.0 x 103 s          

                                          Pm = W/t = (5.0 x 107)/ (4.0 x 103) = 1.25 x 104 W = 12.5 kW

 Solution 2 : Pm = F v =( 500)(25) = 1.25 x 104 W = 12.5 kW

 

Remark

Because the velocity remains constant, the kinetic energy does not change 

All energy is used to be overcome the fricional forces 

Chemical energy  is converted into (frictional) heat