1.2 Justificaci´ on e importancia
3.1.3 Proceso de hidrataci´ on del cemento Portland
How much work does it do in 1 minute? [600 kJ] 2. Determine the power required to lift a load through a height of 20 m in 12.5 s if the force required is 2.5 kN. [4 kW] 3. 25 kJ of work is done by a force in mov- ing an object uniformly through 50 m in 40 s. Calculate (a) the value of the force, and (b) the power.
4. A car towing another at 54 km/h exerts a steady pull of 800 N. Determine (a) the work done in 14 hr, and (b) the power required.
[(a) 10.8 MJ (b) 12 kW] 5. To what height will a mass of weight 500 N be raised in 20 s by a motor using
4 kW of power? [160 m]
6. The output power of a motor is 10 kW. Determine (a) the work done by the motor in 2 hours, and (b) the energy used by the motor if it is 72% efficient. [(a) 72 MJ (b) 100 MJ] 7. A car is travelling at a constant speed of 81 km/h. The frictional resistance to motion is 0.60 kN. Determine the power required to keep the car moving at this
speed. [13.5 kW]
8. A constant force of 2.0 kN is required to move the table of a shaping machine when a cut is being made. Determine the power required if the stroke of 1.2 m is
completed in 5.0 s. [480 W]
9. A body of mass 15 kg has its speed reduced from 30 km/h to 18 km/h in 4.0 s. Calculate the power required to effect this change of speed. [83.33 W] 10. The variation of force with distance for a vehicle that is decelerating is as follows: Distance (m) 600 500 400 300 200 100 0 Force (kN) 24 20 16 12 8 4 0 If the vehicle covers the 600 m in 1.2 minutes, find the power needed to bring the vehicle to rest. [100 kW] 11. A cylindrical bar of steel is turned in a lathe. The tangential cutting force on the tool is 0.5 kN and the cutting speed is 180 mm/s. Determine the power absorbed in cutting the steel. [90 W]
14.4
Potential and kinetic energy
Mechanical engineering is concerned principally with two kinds of energy, potential energy and kinetic energy.
Potential energyis energy due to the position of the body. The force exerted on a mass of mkg is mg N (whereg=9.81 m/s2, the acceleration due to gravity). When the mass is lifted vertically through a height h m above some datum level, the work done is given by: force×distance=(mg (h)J. This work done is stored as potential energy in the mass. Hence,
potential energy=mgh joules
(the potential energy at the datum level being taken as zero).
Kinetic energy is the energy due to the motion of a body. Suppose a forceF acts on an object of massmoriginally at rest (i.e.u=0) and accelerates it to a velocityv in a distances:
work done=force×distance=F s
=(ma) (s) (if no energy is lost) wherea is the acceleration
Since v2 =u2+2as (see Chapter 11) andu =0, v2=2as, from which
a= v 2 2s, hence,
work done=(ma)(s)
=(m) v 2 2s (s)=1 2mv 2
This energy is called the kinetic energy of the mass m, i.e.
kinetic energy= 12mv2 joules
As stated in Section 14.2, energy may be converted from one form to another. The principle of con- servation of energy states that the total amount of energy remains the same in such conversions, i.e. energy cannot be created or destroyed.
In mechanics, the potential energy possessed by a body is frequently converted into kinetic energy, and vice versa. When a mass is falling freely, its potential energy decreases as it loses height, and its kinetic energy increases as its velocity increases. Ignoring air frictional losses, at all times:
If friction is present, then work is done overcoming the resistance due to friction and this is dissipated as heat. Then,
Initial energy=final energy
+work done overcoming frictional resistance Kinetic energy is not always conserved in collisions. Collisions in which kinetic energy is conserved (i.e. stays the same) are called elastic collisions, and those in which it is not conserved are termed inelastic collisions.
Problem 22. A car of mass 800 kg is climbing an incline at 10°to the horizontal. Determine the increase in potential energy of the car as it moves a distance of 50 m up the incline.
With reference to Figure 14.10, sin 10°= opposite
hypotenuse = h
50,
from which, h=50 sin 10°=8.682 m.
h 50 m 10° Figure 14.10 Hence, increase in potential energy=mgh =800 kg×9.81 m/s2 ×8.682 m =68140 J or 68.14 kJ
Problem 23. At the instant of striking, a hammer of mass 30 kg has a velocity of 15 m/s. Determine the kinetic energy in the hammer.
Kinetic energy= 12mv2= 12(30 kg)(15 m/s)2 i.e.
kinetic energy in hammer=3375 J or3.375 kJ
Problem 24. A lorry having a mass of 1.5 t is travelling along a level road at 72 km/h. When the brakes are applied, the speed decreases to 18 km/h. Determine how much the kinetic energy of the lorry is reduced.
Initial velocity of lorry, v1 =72 km/h =72km h ×1000 m km× 1 h 3600 s = 72 3.6=20 m/s, final velocity of lorry,
v2 = 18
3.6=5 m/s and mass of lorry, m=1.5t=1500 kg
Initial kinetic energy of the lorry = 1
2mv 2
1 = 12(1500)(20)
2 =300 kJ Final kinetic energy of the lorry
= 1 2mv
2
2 = 12(1500)(5)
2=18.75 kJ Hence,the change in
kinetic energy=300−18.75=281.25 kJ (Part of this reduction in kinetic energy is converted into heat energy in the brakes of the lorry and is hence dissipated in overcoming frictional forces and air friction).
Problem 25. A canister containing a meteorology balloon of mass 4 kg is fired vertically upwards from a gun with an initial velocity of 400 m/s. Neglecting the air resistance, calculate (a) its initial kinetic energy, (b) its velocity at a height of 1 km, (c) the maximum height reached.
(a) Initial kinetic energy= 12mv2
= 1
2(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+12mv2 =320 000 from which, 12(4)v2 =320 000−39 240 =2 80 760 Hence v= 2×2 80 760 4 =374.7 m/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= 1
2mv 2, from which, velocity v=2gh
=(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= 12mv2 = 21(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 = 21mv2= 12(700)v2
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= 4.34×10 3 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.
Determine the braking power required to give this change of speed.
Change in kinetic energy of car = 1 2mv 2 1−12mv 2 2, where m=mass of car=600 kg,
v1=initial velocity=90 km/h
= 90
3.6 m/s=25 m/s, and v2=final velocity=54 km/h
= 54 3.6 m/s=15 m/s. Hence, change in kinetic energy= 12 m(v21−v22) = 1 2(600)(25 2−152) =120 000 J.
Braking power= change in energy time taken = 120 000 J
15 s
=8000 W or 8 kW Now try the following exercises
Exercise 71 Further problems on poten-