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Modified Work Energy Equation

Equation (6.4) can be modified as

 

T1 +Vg1 + V e1 +U1-2 = T2 +Vg2 +v e2

or                                               U1-2 = ∆(T +Vg +V e) = ∆E

 

where                                               E = total mechanical energy

 

If the total mechanical energy E is constant

Then                                        ∆E = 0  ;  U1-2 = 0

 

E = constant, expresses the law of conservation of dynamical energy.

 

 

SOLVED EXAMPLES

 

Example 6.1. A block of steel of weight 8000 N is placed on a smooth inclined plane which makes an angle 30° with the horizontal. Calculate the work done in pulling the block up for a length of 500 mm.

 

Solution .                N1 =8000 cos 300 (normal reaction)

P =8000 sin 300 (for impending stage)

= 4000 N.

Work done in pulling the block up

= p x 500 /1000 = p /2 = 4000 /2 = 2000 N-m

= 2000 joul . Ans

 

Example 6.2. In problem 6.1, if the inclined plane is rough having 11 = 0.3 (between block and plane), then find the work done.

 

Solution.                              N1 = 8000 cos 30°

For equilibrium,           P = 8000 sin 30° + µN1

Frictional force ,          µN1 = 0.3 x 8000 cos 30°

= 2078.46 N.

P =8000 sin 30° + 2078.46 = 6078.46 N.

Work done = p x 500 /1000 = p /2 = 6078.46 /2 = 3039.23 N-m.Ans
 

Example 6.3. Calculate the work done by electric motor in winding up a uniform cable which hangs from a hoisting drum if its free length is 20 m and weighs 1000 N. The drum is rotated by the motor.

 Solution. The weight of the cable per metre

= 1000 / 20 = 50 N

Weight of elemental length dy

= 50 dy.

The work done            dW = y dy x 50

=50 y dy

Example 6.4. A goods train weighing 4000 KN is pulled by an engine on a level track at a constant speed of 1 0 m/s. The resistance due to friction is 5 N per kN of the train’s weight.

Calculate the power of the engine.

 

Solution. Frictional Force= S x 4000 = 20000 N

P = 20000 N, P = Force developed by train engine

Power= Pv = 20000 x 10 20000 N

= 200000 N-m/s = 2 x 105 watt. Ans.

 

Example 6.5. In problem 6.4, if the train moves with an acceleration of 0.5 m/s2 on the level track after attaining a speed of 10 m/s, then find the power of the engine.

 

Solution.      Net Force= mass x acceleration

P- 20000 = ( 4000 x (10)3 / 9.81 ) x 0.5

P = 223873.59 N

Power = pv

= 223873.59 x 10 N –m /s

= 2238735 J /s  = 2238.7359 kW