Mechanical Engineering Principles

164 MECHANICAL ENGINEERING PRINCIPLES

(a) Initial kinetic energy=^12 mv^2

=^12 ( 4 )( 400 )^2 =320 kJ

(b) At a height of 1 km, potential energy =
mgh= 4 × 9. 81 × 1000 = 39 .24 kJ. By the
principle of conservation of energy: potential
energy+ kinetic energy at 1 km = initial
kinetic energy.

Hence 39 240+^12 mv^2 =320 000

from which,^12 ( 4 )v^2 =320 000−39 240

=2 80 760

Hence v=

√(

2 ×2 80 760

4

)

= 374 .7m/s

i.e.the velocity of the canister at a height of

1 km is 374.7 m/s

(c) At the maximum height, the velocity of the

canister is zero and all the kinetic energy has

been converted into potential energy. Hence,

potential energy = initial kinetic energy =

3 20 000 J(from part (a))Then,

320000 =mgh=( 4 )( 9. 81 )(h),

from which, heighth=

3 20 000

( 4 )( 9. 81 )

=8155 m

i.e.the maximum height reached is 8155 m.

Problem 26. A piledriver of mass 500 kg

falls freely through a height of 1.5 m on to a

pile of mass 200 kg. Determine the velocity

with which the driver hits the pile. If, at

impact, 3 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

200 mm, determine (a) the common velocity

immediately after impact, (b) the average

resistance of the ground.

The potential energy of the piledriver is converted

into kinetic energy.

Thus potential energy=kinetic energy,

i.e. mgh=^12 mv^2 ,

from which, velocityv=

√

2 gh

=

√

( 2 )( 9. 81 )( 1. 5 )

= 5 .42 m/s.

Hence,the piledriver hits the pile at a velocity of

5.42 m/s.

(a) Before impact, kinetic energy of

pile driver=^12 mv^2 =^12 ( 500 )( 5. 42 )^2

= 7 .34 kJ

Kinetic energy after impact = 7. 34 − 3 =

4 .34 kJ. Thus the piledriver and pile together

have a mass of 500 + 200 = 700 kg and possess

kinetic energy of 4.34 kJ.

Hence 4. 34 × 103 =^12 mv^2 =^12 ( 700 )v^2

from which, velocityv=

√

√

√

√

(

2 × 4. 34 × 103

700

)

= 3 .52 m/s

Thus,the common velocity after impact is

3.52 m/s.

(b) The kinetic energy after impact is absorbed in

overcoming the resistance of the ground, in a

distance of 200 mm.

Kinetic energy=work done

=resistance×distance

i.e. 4. 34 × 103 =resistance× 0. 200 ,

from which,

resistance=

4. 34 × 103

0. 200

=21700 N

Hence,the average resistance of the ground

is 21.7 kN.

Problem 27. A car of mass 600 kg reduces

speed from 90 km/h to 54 km/h in 15 s.