1200 kg ( 9.8 m/s^2)

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Transcript 1200 kg ( 9.8 m/s^2)

• effect of force on the displacement of the
• can be computed by multiplying the force
by the parallel displacement
• force X displacement (assuming that they
are parallel)
• SI base unit is joule ( J).
W [Joule] = ( F cos q ) d
A bag is pulled with an angle of 60° with the
horizontal. The work done on the bag is
300 J. What is the displacement of the
bag if the force applied is 30 N?
300J = 30 N (cos 60°)( d)
d = 20 m
Work Done Against Gravity
How much work should
be done in lifting a
200-kg load of bricks
to a height of 10 m on
a construction
Formula : Work to lift an object = mgh
W = 200 kg ( 9.8 m/s^2)( 10 m)
= 19, 600 N-m or 19,600 J
Remember that…
• In the absence of friction and other
external forces, the work done in lifting an
object with a mass m to a height of h is
mgh regardless of the exact path taken.
The particular route taken by an object
being raised is not significant.
• The work done to go
up using the back
portion is the same as
the work done to go
up at the same height
using the steps.
• the rate of doing work
power = work done/ time
The SI unit for power is
watt ( W).
Units for Power
• There are other units for power.
• The conversion factors for the units for
power are
1 hp (horsepower) = 746 W
1 kW = 1000 W
Kilowatt-hour (kWh) is a unit for work.
An electric motor with an output of 15 kW
provides power for the elevator of a
building. If the total mass of the loaded
elevator is 1000 kg, how long will it take
for it to rise by 30 m from the ground floor
to the top floor?
Power = mgh = 1000kg(9.8m/s^2)(30m)
Since power is 15, 000 W then t is 19.6 s.
Efficiency (Eff) = Power output X 100
Power input
Sample Problem
A crane whose motor has Solution:
a power input of 4.0 kW Compute for power output:
raises a 1200 –kg beam
1200 kg ( 9.8 m/s^2) 30m
through a height of 30 m
90 s
in 90 s. Find the
= 3 920 W or 3.92 kW
Eff = 3.92 kW x 100
4.0 kW
= 98 %
Gravitational Potential Energy
• For example, a 10- kg
object was lifted 1m
from the ground.
GPE = 10kg( 9.8 m/s2) ( 1m)
= 98 J.
Kinetic Energy (KE)
• For example , a 10-kg
object is moving at
2m/s to the east.
• KE = 10 kg( 2m/s)2
= 20 J
Total Mechanical Energy
• sum of the potential energy and kinetic
energy of an isolated system
Law of Conservation of Energy.
The energy is conserved in a given system,
although energy transformations from one
form to another may occur.
Conservation of Mechanical Energy
The mechanical energy of an isolated
system is conserved.
KE1 + PE1 =
KE2 + PE2
A 0.2 – kg ball is tied to a string with a length
of 1.5 m. The other end of the string is tied to a
rigid support. The ball is held straight out
horizontally from the point of support, with the
string taught and is then released.
What is the maximum potential energy that the
ball can have (with respect to its lowest point)?
PE = mgh
= 0.2 kg (9.8 m/s^2)( 1.5m)
= 2.94 J
The PE can transform to
kinetic energy while the bob
is moving downward.
1.5 m
Linear Momentum
• vector quantity that describes the
tendency of an object to continue
moving at constant velocity
• product of mass and velocity
• denoted by the symbol ‘p’
p = mv
• the SI unit for momentum is kg m/s
• vector quantity that describes the change
in momentum due to the application of
force in a period of time
• product of force and time when the force is
• expressed in N s
FΔt = mΔv
• Impulse is equal to the change in
• For example there is a change in
momentum from 2 kg m/s to 20 kg m/s.
• Therefore the impulse is 18 kgm/s.
• If the contact time is 0.1 s then the force
applied is 180 N since the impulse is
Law of Conservation of Momentum
“The total linear momentum of the system is
conserved if there is no external force
acting on the system .”
total momentum
before collision
total momentum
after collision
mA viA + mBviB = mAvfA + mBvfB
Three Types of Collision
Elastic Collision - KE is conserved
Inelastic Collision – KE is not conserved
and usually lost to heat and sound
Perfectly Inelastic Collision – objects stick
together after the collision and KE loss is