Transcript Work, Energy and Power

Work, Energy and Power
Work, Energy and Power
• Work = Force component x displacement
• Work = Fxx
• When the displacement is perpendicular to
the force, no work is done.
• When there is no displacement, no work is
done
Example 8.1
• What work is done by a 60-N force pulling
the wagon below a distance of 50 meters
when the force transmitted by the handle
makes an angle of 30o with the horizontal?
Example 8.1
• What work is done by a 60-N force pulling
the wagon below a distance of 50 meters
when the force transmitted by the handle
makes an angle of 30o with the horizontal?
• Work =(Fcosθ)x = (60 N)(cos30o)(50m)
• Work = 2600 Nm
Resultant Work
• When work involves several forces we can
calculate the resultant work 2 ways
• 1st – Find the work done by each force and
add them
• 2nd – Find the resultant force and multiply
by the displacement
Example 8.2
• A push of 80 N moves a 5 kg block up a 30o inclined plane, as
shown below. The coefficient of kinetic friction is 0.25, and the
length of the plane is 20 m.
• (a) Compute the work done by each force acting on the block
• (b)Show that the net work done by these forces is the same as the
work done by the resultant force
Example 8.2
• Consider the work done by each force
• Work done by the normal force n is 0
• Work done by the pushing force P is (80 N)(1)(20 m) =
1600 J
• The friction force is μkN = -10.6 N, the negative sign
indicates this force opposes the motion. The work done
by the friction force is (-10.6 N)(20 m) = -212 J
• Work done by x component of the weight =
-(24.5 N)(20 m) = -490 N
• Net work = 0 + 1600 J – 212 J -490 J = 898 J
Example 8.2
• Determine the resultant force and
calculate the work done by the resultant
force
• Fr = P –Ff –Wx
• =80N – 10.6N -24.5N = 44.9N
• Net work is
• (44.9N)(20 m) = 898 J
Energy
• Energy may be thought of as anything that can
be converted to work.
• We will consider two kinds of energy
• Kinetic Energy – Energy possessed by a body
by virtue of its motion
• KE = ½ mv2
• Potential Energy – Energy possessed by a
system by virtue of its position or condition.
• Gravitational Potential Energy, U = mgh
Kinetic Energy
• Consider the following examples of objects
with kinetic energy:
•
•
•
•
1200 kg car, v = 80 km/h, (22.2 m/s)
KE = ½ (1200 kg)(22.2 m/s)2 = 296,000 J
20 g bullet v = 400 m/s
KE = ½ (0.02 kg)(400 m/s)2 = 1,600 J
Example 8.3
• Find the kinetic energy of a 4 kg
sledgehammer with a velocity of 24 m/s.
• K = ½ mv2 = ½(4 kg)(24 m/s)2
• K = 1150 J
Work-Energy Theorem
• When work is done on a mass to change
its motion, The work done is equal to the
change in kinetic energy
• Fx = 1/2mvf2 – 1/2mvi2
Example 8.5
• What average force is necessary to stop a
16 gram bullet traveling at 260 m/s as it
penetrates 12 cm into a block of wood?
• Fx = 1/2m(0)2 – ½ m(vf)2 or Fx = -1/2mvf2
• F = -mvf2/2x = -4510 N
Potential Energy
• The energy that systems possess by virtue
of their positions or conditions is potential
energy.
• When work is done against the force of
gravity, it gives the object gravitational
potential energy
• Gravitational potential energy is found
using U = mgh
Potential Energy
• Lifting a mass to a height h requires work
mgh. If the mass falls it acquires kinetic
energy equal to its potential energy
Example 8.6
• A 1.2 kg toolbox is held 2 meters above the top of a table
that is 80 cm from the floor. Find the potential energy
relative to the top of the table and relative to the floor.
• The potential energy relative to the tabletop is:
• U =mgh =(1.2 kg)(9.8 m/s2)(2 m) = 23.5 J
• The potential energy relative to the floor is
• U = mgh= (1.2 kg)(9.8 m/s2)(2.0 +0.8 m) = 32.9 J
Conservation of Energy
• In the absence of air resistance or other
dissipative forces, the sum of the potential
and kinetic energies is a constant.
Provided that no energy is added to the
system.
Conservation of Energy
As the ball the falls from height h, the sum of
potential, U and Kinetic, K energies is
conserved.
Example 8.8
• In the figure below, a 40 kg wrecking ball is pulled to one
side until it is 1.6 m above its lowest point. Neglecting
friction, what will be its velocity as it passes through its
lowest point?
Example 8.8
• Applying Conservation of Energy
• mgh + 0 = 0 + ½ mv2
• mgh = ½ mv2
v  2 gh  2(9.8m/s )(1.6m
2
• V = 5.6 m/s
Energy and Friction Forces
• When dissipative forces are present, they
must be included in the final energy total
• Initial total energy = final total energy +
losses due to friction
Example 8.9
• A 20 kg sled rests at the top of a 30o slope 80 meters in
length, as shown below.
• If µk = 0.2, what is the velocity at the bottom of the
incline?
Example 8.9
• Using the Conservation of Energy
equation, we get
• mgh0 + ½ mvo2 = mghf + ½ mvf2 +Ffx
• Using vo = 0, and Ff = µkN= (0.2)(170 N)
• We get 7840 J = ½ (20 Kg)vf2 + 2720 J
• vf = 22.6 m/s
Power
• Power is the rate at which work is
accomplished.
• P = work/time
• The unit of power is the watt, 1kilowatt =
1000 watts, 1 hp =746 watts
Example 8.10
• A loaded elevator has a total mass of 2800
kg and is lifted to a height of 200 m in a
time of 45 seconds. Calculate the average
power.
• P = Fx/t = mgh/t
• P = (2800 kg)(9.8 m/s2)(200 m)/45 s
• P = 1.22 x 105 W = 122 kW