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

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).

WORK

W [Joule] = ( F cos ) 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 building?

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

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.

POWER

• 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?

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

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

Efficiency

Efficiency (Eff) = Power output X 100

Power input

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 %

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

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

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 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

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

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

• 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.

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

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