Month: March 2013

Efficiency

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Efficiency

The efficiency of a device is defined as the percentage of the energy input that is transformed into useful energy.

Example:
In the example above, the input power is 100J/s, the desire output power (useful energy) is only 75J/s, the remaining power is lost as undisire output. Therefore, the efficiency of this machine is

Efficiency= 75 100 ×100%=75%

Air Conditioner


  1. Switch off the air conditioner when not in use.
  2. Buy the air conditioner with suitable capacity according to the room size.
  3. Close all the doors and windows of the room to avoid the cool air in the room from flowing out.

Refrigerator


  1. Always remember to close the door of refrigerator.
  2. Open the refrigerator only when necessarily.
  3. Always keep the cooling coil clean.
  4. Defrost the refrigerator regularly.
  5. Choose the refrigerator with capacity suitable for the family size.
  6. Refrigerator of large capacity is more efficient compare with refirgerator of small capacity.

Lamp or Light Bulb


  1. Use fluorecent bulb rather than incandescent bulb. Fluorescent bulbs are much more efficient than incandescent bulbs.
  2. Use a lamp with reflector so that more light is directed towards thr desirable place.

Washing Machine


  1. Use front-loading washing machine rather than top-loading wahing machine because it uses less water and electricity.
  2. Use washing machine only when you have sufficient clothes to be washed. Try to avoid washing small amount of clothes.

Power

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Power

Power is the rate at which work is done, which means how fast a work is done.

Formula:

Example:
An electric motor takes 20 s to lift a box of mass 20kg to a height of 1.5 m. Find the amount of work done by the machine and hence find the power of the electric motor.

Answer:
Work done,
W = mgh
W = (20)(10)(1.5) = 300J

Power,

Relationship between Energy and Work Done

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During a conversing of energy,

Amount of Work Done = Amount of Energy Converted

Example:
A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and it stops at B after t seconds. What is the work done to overcome friction?

Answer:
In this case, kinetic energy is converted into heat energy due to the friction. The work done to overcome the friction is equal to the amount of kinetic energy converted into heat energy, hence

Potential Energy

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Energy

  1. Energy is defined as the capacity to do work.
  2. Work is done when energy is converted from one form to another.
  3. The unit of work is Nm or Joule(J)

 

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object as the result of its vertical position (i.e., height).

Formula:

 

Example:
A ball of 1kg mass is droppped from a height of 4m. What is the maximum kinetic energy possessed by the ball before it reached the ground?

Answer:
According to the principle of conservation of energy, the amount of potential energy losses is equal to the amount of kinetic energy gain.

Maximum kinetic energy
= Maximum potentila energy losses
= mgh = (1)(10)(4) = 40J

 

Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing.

Formula:

Example:

The diagram above shows a spring with a load of mass 0.5kg. The extension of the spring is 6cm, find the energy stored in the spring.

Answer:
The energy stored in the spring is the elastic potential energy.

 

Finding Work Done from a Graph

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Finding Work from Force-Displacement Graph
In a Force-Displacement graph, work done is equal to the area in between the graph and the horizontal axis.

Example:

The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m. Find the work done by the force to move the trolley.

Answer:
In a Force-Displacement graph, work done is equal to the area below the graph. Therefore, work done
W= 1 2 (10)(8)
W = 40Nm
W = 40J 

Work Done by/Against the Gravity

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Work Done Against the Force of Gravity

Example:
Ranjit runs up a staircase of 35 steps. Each step is 15cm in height. Given that Ranjit’s mass is 45kg, find the work done by Ranjit to reach the top of the staircase.

Answer:
In this case, Ranjit does work to overcome the gravity.

Ranjit’s mass = 45kg
Vertical height of the motion, h = 35 × 0.15
Gravitational field strength, g = 10 ms-2

Work done, W = ?
W = mgh
W = (45)(10)(35 × 0.15)
W = 2362.5J

Work

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Work

  1. Work done by a constant force is given by the product of the force and the distance moved in the direction of the force.
  2. The unit of Nm(Newton metre) or J(Joule).
  3. Work is a scalar quantity.

Formula


When the direction of force and motion are same, θ = 0o, therefore cosθ = 1
Work done,

W = F × s

Example:

A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. Calculate the work being done by the force.

Answer
Work done,
W = F × s × cos θ
W = 50 × 3.0 × cos30o = 129.9J

Example:

The diagram above shows a 10N force is pulling a metal. The friction between the block and the floor is 5N. If the distance travelled by the metal block is 2m, find

  1. the work done by the pulling force
  2. the work done by the frictional force

Answer:
(a) The force is in the same direction as the motion. Work done by the pulling force,
W = F × s
W = (10)(2) = 20J

(b) The force is not in the same direction of motion, work done by the frictional force
W = F × s × cos180o
W = (5)(2)(-1) = -10J

Pulley

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  1. There are 2 types of pulley, the fixed pulley and the movable pulley.

(“Fixed pulley” by César Rincón available under a Creative Commons Attribution-Share Alike 3.0 Unported)
(“Movable pulley” by Johjak available under a Creative Commons Attribution-Share Alike 3.0 Unported)
  1. In SPM, we will only discuss the fixed pulley.
  2. The fixed pulley change the direction, without changing the magnitude of the force.
  3. In SPM, we assume all pulleys are smooth (no friction) unless it is stated otherwise.

Apparent Weight of an Object in a Lift

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  1. When a man standing inside an elevator, there are two forces acting on him.
    1. His weight, (W) which acting downward.
    2. Normal reaction (R), acting in the opposite direction of weight.
  2. The reading of the balance is equal to the normal reaction (R).
  3. Figure below shows the formula to calculate the reading of the balance at different situation.

Example 1:
Subra is standing on a balance inside an elevator. If Subra’s mass is 63kg, find the reading of the balance when the lift,

  1. stationary
  2. moving upward with a constant velocity, 15 ms-1.,
  3. moving upward with a constant acceleration, 1 ms-2.
  4. moving downward with a constant acceleration, 2 ms-2.

Answer:
a.

W = mg
W = (63)(10) = 630N

b.

W = mg
W = (63)(10) = 630N

c.

R = mg+ma
R = (63)(10)+(63)(1)
R = 693N

d.

R = mg−ma
R = (63)(10)−(63)(2)
R = 504N

 

Example 2:
A 54kg boy is standing in an elevator. Find the force on the boy’s feet when the elevator

  1. stands still
  2. moves downward at a constant velocity of 3 m/s
  3. decelerates downward with at 4.0 m/s2,
  4. decelerates upward withat 2.0 m/s2.

Answer:
a.

W = mg
W = (54)(10) = 540N

b.

W = mg
W = (54)(10) = 540N

c.

R=mg+ma
R=(54)(10)+(54)(4)
R=756N

d.

R=mg−ma
R=(54)(10)−(54)(2)
R=432N