Mechanical Engineering Principles

WORK, ENERGY AND POWER 165

Determine the braking power required to

give this change of speed.

Change in kinetic energy of car

=^12 mv^21 −^12 mv^22 ,

where m=mass of car=600 kg,

v 1 =initial velocity=90 km/h

=

90

3. 6

m/s=25 m/s,

and v 2 =final velocity=54 km/h

=

54

3. 6

m/s=15 m/s.

Hence, change in

kinetic energy=^12 m(v^21 −v^22 )

=^12 ( 600 )( 252 − 152 )

=120 000 J.

Braking power=

change in energy

time taken

=

120 000 J

15 s

=8000 W or 8kW

Now try the following exercises

Exercise 71 Further problems on poten-

tial and kinetic energy

(Assume the acceleration due to gravity,

g= 9 .81 m/s^2 )

1. An object of mass 400 g is thrown verti-
   cally upwards and its maximum increase
   in potential energy is 32.6 J. Determine
   the maximum height reached, neglecting
   air resistance. [8.31 m]


2. A ball bearing of mass 100 g rolls down
   from the top of a chute of length 400 m
   inclined at an angle of 30°to the horizon-
   tal. Determine the decrease in potential
   energy of the ball bearing as it reaches
   the bottom of the chute. [196.2 J]


3. A vehicle of mass 800 kg is travelling at
   54 km/h when its brakes are applied. Find
   the kinetic energy lost when the car comes
   to rest. [90 kJ]
   4. Supplies of mass 300 kg are dropped
   from a helicopter flying at an altitude of
   60 m. Determine the potential energy of
   the supplies relative to the ground at the
   instant of release, and its kinetic energy
   as it strikes the ground.
   [176.6 kJ, 176.6 kJ]
   5. A shell of mass 10 kg is fired verti-
   cally upwards with an initial velocity
   of 200 m/s. Determine its initial kinetic
   energy and the maximum height reached,
   correct to the nearest metre, neglecting air
   resistance. [200 kJ, 2039 m]
   6. The potential energy of a mass is
   increased by 20.0 kJ when it is lifted
   vertically through a height of 25.0 m. It
   is now released and allowed to fall freely.
   Neglecting air resistance, find its kinetic
   energy and its velocity after it has fallen
   10.0 m. [8 kJ, 14.0 m/s]
   7. A piledriver of mass 400 kg falls freely
   through a height of 1.2 m on to a pile
   of mass 150 kg. Determine the velocity
   with which the driver hits the pile. If,
   at impact, 2.5 kJ of energy are lost due
   to heat and sound, the remaining energy
   being possessed by the pile and driver as
   they are driven together into the ground
   a distance of 150 mm, determine (a) the
   common velocity after impact, (b) the
   average resistance of the ground.
   [4.85 m/s (a) 2.83 m/s (b) 14.68 kN]

14.5 Kinetic energy of rotation

Whenlinear motiontakes place,

kinetic energy=

∑m

2

v^2 ,

but whenrotational motiontakes place,

kinetic energy=

1

2

∑

m(ωr)^2

Sinceωis a constant,

kinetic energy=ω^2

1

2

∑

mr^2