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Presentation on theme: "2.10 : WORK, ENERGY, POWER AND EFFICIENCY"— Presentation transcript:


2 WORK Is defined as the product of the applied force and the distance moved in the direction of the force. Work = force x distance W = F x s SI unit is the Joule (J) Is a scalar quantity

3 No work done when: A force is applied but no displacement occurs.
An object undergoes a displacement with no applied force acting on it. The direction of motion is perpendicular to the applied force.

Direction of force, F is same as the direction of displacement, s. W = F x s F s F

5 Direction of force, F is not same as the direction of displacement, s.
W = F cos θ x s W = F s cos θ F s F θ θ F cos θ

6 Direction of force, F is perpendicular to the direction of displacement, s.
W = F x s W = F cos 90° x s W = 0 F s F

7 Example 1 If a box is pushed with a force of 40 N and it moves steadily through a distance of 3m in the direction of the force, calculate the work done. Answer: W = 120 J

8 Example 2 A women pulls a suitcase with a force of 25 N at an angle of 60° with the horizontal. What is the work done by the women if the suitcase moves a distance of 8 m along the floor? Answer,: W = 100 J

9 FORCE–DISTANCE GRAPH Area under a force–distance graph = work done.
Work = F x s = a x b Force, F a Distance, s b

An upward force, F is applied to lift the object of weight, W to a height, h. W = F x h W = mgh F h W

11 ENERGY Is defined as potential or the ability to do work. Form energy:
Gravitational potential energy Kinetic energy Heat energy Sound energy Electrical energy SI unit in Joule, J and it scalar quantity. Energy is transferred from 1 object to another when work is done.

12 KINETIC ENERGY, Ek Is a energy possessed by a moving object.
It scalar quantity. SI unit in Joules, J. Formula : Ek = 1 mv2 2 Ek = 1(mv2) – 1(mu2) Object moving from initial velocity to final velocity

13 POTENTIAL ENERGY, Ep Defined as energy of an object due to its higher position in the gravitational field. Depend on mass, gravitational field and height. Formula: Ep = mgh # m = mass # g = acceleration due to gravity # h = difference between height

14 Example 3 In a school sports event, a student of mass 40 kg runs past the finishing line with a velocity of 7 ms-1. Calculate his kinetic energy. Answer: Ek = 980 J

15 Example 4 A durian fruit hanging from its branch has gravitational potential energy due to its higher position above the ground. The mass of the fruit is 2.5 kg and it hangs 3 m above the ground. What is the gravitational potential energy of the fruit? (g = 10 ms-2) Answer: Ep = 75 J

State that energy cannot be created or destroyed. It can be transformed from one form to another. The total energy in a system is constant. This means there is no energy gained or lost in a process. Formula : mgh = 1mv2 2

17 Example: On winning a match, a tennis player hits a tennis ball vertically upward with an initial velocity of 25ms-1. What is the maximum height attained by the ball? (g = 10ms-2) Answer ; h = m

18 POWER Is the rate at which work is done or rate at which energy is transformed. Formula; Power, P = work energy transformed time taken time taken P = W @ F x F x Fv t t t P = E t SI unit is watt (W) or Js-1 Is scalar quantity P ∞ W if t constant P ∞ 1 / t if work constant

19 Example: A weightlifter lifts 160kg of weights from the floor to a height of 2m above his head in a time of 0.8s. What is the power generated by the weightlifter during this time? (g = 10ms-2) Answer : P = 4000 W

20 EFFICIENCY The percentage of the input energy that is transformed to useful form of output energy. Formula : efficiency = useful energy output x 100% energy input = Eo x 100% Ei

21 Also can be calculated in terms of power.
efficiency = useful power output x 100% power input = Po x 100% Pi

22 Example: An electric motor in a toy crane can lift a 0.12 kg weight through a height of 0.4m in 5s. During this time, the batteries supply 0.80 J of energy to the motor. Calculate The useful energy output of the motor The efficiency of the motor Answer: Eo = 0.48J Efficiency = 60 %

23 Exercise A steel ball of mass 2 kg is released from a height of 8 m from the ground. On hitting the ground, the ball rebounds to a height of 3.2 m as shown in figure. If air resistance can be neglected and the acceleration due to gravity g = 10 ms-2, find The kinetic energy of the ball before it reaches the ground. The velocity of the ball on reaching the ground. The kinetic energy of the ball as it leaves the ground on rebound. The velocity of the ball on rebound. 8 m 3.2 m Ek =160 J v1 = ms-1 Ek = 64 J v2 = 8 ms-1

24 Exercise A car moves at a constant velocity of 72 kmj-1. Find the power generated by the car if the force of friction that acts on it is 1500 N. Answer : P = W


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