work effect of force on the displacement of the object can be computed by multiplying the force by...

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Work • effect of force on the displacement of the object • 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).

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Page 1: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Work

• effect of force on the displacement of the object

• 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).

Page 2: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

WORK

W [Joule] = ( F cos ) d

Page 3: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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

Page 4: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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 building?

Page 5: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Solution:

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

Page 6: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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.

Page 7: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

• 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.

Page 8: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

POWER

• the rate of doing work

power = work done/ time

The SI unit for power is watt ( W).

Page 9: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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.

Page 10: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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?

Solution: Power = mgh = 1000kg(9.8m/s^2)(30m) t t

Since power is 15, 000 W then t is 19.6 s.

Page 11: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Efficiency

Efficiency (Eff) = Power output X 100

Power input

Page 12: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Sample Problem

A crane whose motor has a power input of 4.0 kW raises a 1200 –kg beam through a height of 30 m in 90 s. Find the efficiency.

Solution:Compute for power output:

1200 kg ( 9.8 m/s^2) 30m

90 s

= 3 920 W or 3.92 kW

Eff = 3.92 kW x 100

4.0 kW

= 98 %

Page 13: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Gravitational Potential Energy (GPE)

• For example, a 10- kg object was lifted 1m

from the ground.

GPE = 10kg( 9.8 m/s2) ( 1m)

= 98 J.

h

m

v

Page 14: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Kinetic Energy (KE)

• For example , a 10-kg object is moving at 2m/s to the east.

• KE = 10 kg( 2m/s)2

2

= 20 J

Page 15: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Total Mechanical Energy

• sum of the potential energy and kinetic energy of an isolated system

Page 16: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Law of Conservation of Energy.

The energy is conserved in a given system, although energy transformations from one form to another may occur.

Page 17: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Conservation of Mechanical Energy

The mechanical energy of an isolated system is conserved.

KE1 + PE1 = KE2 + PE2

Page 18: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

A 0.2 – kg ball is tied to a string with a lengthof 1.5 m. The other end of the string is tied to arigid 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 theball 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

Page 19: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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

Page 20: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

Impulse

• 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 applied

• expressed in N s

Impulse: FΔt = mΔv

Page 21: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

• Impulse is equal to the change in momentum.

• 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

Ft.

Page 22: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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 = total momentum

before collision after collision

mA viA + mBviB = mAvfA + mBvfB

Page 23: Work effect of force on the displacement of the object can be computed by multiplying the force by the parallel displacement force X displacement (assuming

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 maximum