CBSE ADDA

WORK AND ENERGY   9th physics Term-II 1. When a body falls freely towards the earth, then its total energy (a) increases (b) decreases (c) remains constant (d) first increases and then decreases 2. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car (a) does not change (b) becomes twice to that of initial (c) becomes 4 times that of initial (d) becomes 16 times that of initial 3 . In case of negative work the angle between the force and displacement is (a) 00 (b) 450 (c) 900 (d ) 1800 4 . An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same (a) acceleration (b) momenta (c) potential energy (d) kinetic energy 5 . A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational forc

CBSE PHYSICS: LAW OF CONSERVATION OF ENERGY(9th physics) : LAW OF CONSERVATION OF ENERGY Energy can neither be created nor destroyed, but it is transformed from one form to another. Alternatively, whenever energy gets transformed, the total energy remains unchanged. Proof – Freely falling body It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant. Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig. At A, Kinetic energy kE = 0 Potential energy Ep = mgh Total energy E = Ep + Ek = mgh + 0= mgh During the fall, the body is at a position B. The body has moved a distance x from A. At B, velocity v 2 = u 2 + 2as applying, v 2 = 0 + 2ax = 2ax Kinetic energy Ek = 1/2 m v 2 = 1/2 m x 2gx = mgx Potential energy Ep = mg (h – x) Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh If the body reaches the position C. At C, Potential energy Ep = 0 V

Multiple type questions 1).what type of wave is sound wave? (a) radio waves (b) electro magnetic wave (c) mechanical wave (d)transverse wave 2.SONAR works on principle of (a) echo (b)resonance(c)particle nature of wave (d) none 3. Loudness depends on (a) wave length (b) frequency(c) amplitude (d)intensity 4. Pitch increases with increase in (a) energy of sound (b) frequency (c) wave length (d) amplitude 5. speed of sound greatest in (a) milk (b) wood(c) iron (d) air 6. The sound whose frequency is between 20 Hz to 20000 Hz is (a)infrasonic wave (b) audible sound (c)ultrasonic wave (d) radio waves Very Short Answer questions 1. What is echo? 2. what are conditions necessary to hear echo? 3. What is SONAR ? Give its use? 4. How density of mediam affect speed of sound? 5. What is pitch of sound? 6. What is amplitude of sound waves? 7. What is resonance? 8. What is sonometer? 9. Define the terms

Work When a force applied on an object and the object moves in the direction of force, we say that the force has done work on the object. Conditions essential for work to be done are :   1. Some force must act on the object   2. Object move in the direction of force   Hence, the product of the force and the distance moved measures work done.   W = F x S   Where W is the work done, F is the force applied and S is the distance covered by the moving object.  Work is the product of the magnitudes of the force and the displacement, and direction is not taken into account. So work is a  scalar  quantity What is the work done when the force on the object is zero? The work done is zero What would be the work done when the displacement of the object is zero? The work done is zero A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force.  If the force acts on the object all through the displacement, what is the work done in this case? Work do

Class 9 Chapter Work and energy Solved Numerical problems 1. A force of 10N causes a displacement or 2m in a body in its own direction. Calculate the work done by force. 20j Solution: the work done by force = F x S = 10 N x 2m = 20 J 2. How much force is applied on the body when 150 joule of work is done in displacing the body through a distance of 10m in the direction of force?(15 N) Solution: W = F x S Þ F = w/s = 150/10 = 15 N 3. A body of 5kg raised to 2m find the work done(98j) Solution: T he work done to raise a body = PE = mgh = 5kg x 9.8 x 2 = 98 joule 4. A work of 4900j is done on road of mass 50 kg to lift it to a certain height. Calculate the height through which the load is lifted. (10m) Solution: work done on road to lift = mgh Þ 4900 = 50 x 9.8 h Þ h = 10m 5. An engine work 54,000J work by exerting a force of 6000N on it. What is the displacement of the force . (9m) Solution: S = W/F = 54,000J/6000N = 9m 6. A force of 10N acting on a body at an angle of 60 deg. with the

CBSE PHYSICS:Class IX Work Power and Energy notes(Physics) : Work :   In our daily life anything that makes us tired is known as  work . For example, reading, writing, painting, walking, etc. In physics work (W) is said to be done, when a force (F) acts on the body and point  of application of the force is displaced (s) in the direction of force. Work done = force x displacement     W          =     F       x       s (i) If the body is displaced in the same direction of force, Work done is positive Read more »