Skip to main content

9th Work Energy and Power solved Guess Questions

Q. If you want to lift the water to damp against the force of gravity what is the energy used by motor.given that Mass=20Kg ; Height=20m;Gravity=9.8m/s; Time=20min

Ans: Work done by the motor against the gravity, W = mgh = 20×9.8×20 = 3920 J

Power delivered by the motor = W/t =3920/(20×60) = 3.27 J/s

Q. when a ball is lifted to a height of height h when it is drooped just before touching the ground, what would the kinetic energy acting on the ball.

Ans: mgh
According to the conservation of energy Potential energy is converted into kinetic energy just above the ground. Just before touching the ground the kinetic energy of the body is equal to mgh

Q. what are the conditions where the work can be zero?

Ans: Work done is said to be zero if the force is acting at right angles to the displacement of the body or when there is no displacement of the body.
Example: Work done by the force of gravity on box lying on a roof of a bus moving with a constant velocity on a straight road is zero.

Q. An object of mass 1 kg is raised through a height h. Its potential energy increases by 1 Joule. Find the height h.

Ans: Mass:- m = 1 kg, height = h m
PE = mgh, taking g = 10 m/s2
PE = (10h) J,
Now potential energy increases by 1 Joule,
PE' = (10h + 1) Joules
 PE' = mgh' = (10h + 1)
10h' = (10h + 1)
10(h’-h) = 1
(h’-h) = 1/10
if body is on the ground h=0 then, h’ = 0.1m

Q. A rocket of mass 3x10^6 kg taken from a launching pad and acquires vertical velocity of 1 km/s and an altitude if 25m. Calculate its Potential Energy and Kinetic Energy.

Ans: Given: m = 3x10^6 kg, v = 1 km/s or 1000m/s and h = 25m.
Potential Energy = mgh = 3x10^6  kg x 10m/s2 x 25m =7.5x108 J
Kinetic Energy = ½ mv2 = ½ x 3x10^6 x1000m/s x1000m/s = 1.5x 1012J

Q. A car of mass, m = 2000 Kg is lifted up a distance, d = 30m by crane in t = 1 min or 60 sec. Second crane does the same job in t' = 2min or 120 sec. What is the power applied by each crane?
Do the crances consume the same or different amount of fuels? Neglect power dissipation against friction.

Ans: Work done, W=mgh=2000x10x30=600000J
Power, P=W/t=600000J/(60)sec=10000Watt
Similarly, W' = 2000x10x60=600000J
P' = W/t' = 5000Watt
First crane has the more power and consume more fuel.

Q. Why work done is zero in satellite moving in circular path?

Ans: Satellite moving in circular path applying perpendicular force towards the centre.
q = 90 and W= F S Cos q = F S Cos 90 = F S x 0 = 0 J

Q. When a book is lifted up, is there any work done?

Ans: Yes, Positive work by man and negative by the Earth, when a book is lifted up. Book shows displacement in the direction of force. Here the work done is against the gravity.

Q. A person is climbing a palm tree 
i) what is its work done against the force of gravity
ii) What is its work done due to the force applied by the person if the person climbs 10m and the mass of the person is 45kg?

Ans: (i) work done against the force of gravity = the amount of potential energy possessed by the man at height h
(ii) Work done due to the force applied by the person if the person climbs 10m and the mass of the person is 45kg. = m g h =45 x 10 x 10=4500J

Q. (a) State the law of conservation of energy. 
(b) Name the type of energy possessed by stretched slinky, flowing water, stretched rubber band and a speeding car.
(c) An object of mass 50 kg is raised to a height of 600 cm above the ground. What is its potential energy? (g = 10m/s2)

Ans: (a) Law of conservation of energy: Total energy of this universe is conserved (is constant). Energy cannot be created nor be destroyed. But can be transformed from one form to another.
b) A stretched slinky and stretched rubber band possesses elastic potential energy. Flowing water and speeding car possesses kinetic energy.
c) Here, ‘m’ = 50 kg Height from ground, h = 600 cm = 6 m
So, PE = mgh = (50)(10)(6) = 3000 J

Q. A girl having a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4m/s by applying a force. The trolley comes to rest after travelling a distance of 16m. How much work is done on the trolley? How much work is done by the girl?

Ans: Given that: mass of girl = 35 kg ; mass of trolley = 5 kg
Total mass, m = (35 + 5) = 40 kg u = 4 m/s, v = 0, s = 16 m.
Using kinematics equation, v 2 = u 2 + 2as ⇒ 0 = (4)2 + 2a (16)
⇒ 32a = –16 ⇒ a = –0.5 m/s2
Force exerted on trolley, F = ma = 40 × 0.5 = 20 N

(a) Work done on trolley W = FS = (20 N) (16 m) = 320 J

(b) Work done by the girl,
W = FS = (mass of girl) × retardation × S
    = 35 × 0.5 × 16 = 280 J

Q. Define average power?

Ans: The amount of work performed during a period of time is called average power or simply power.

Q. two children A nad B each weighing 20 kg climb a rope upto the height of 10. child A take 10s child B takes 20s to climb. State whether the work done by both is equal or unequal. Who has more power?

Ans: Work done=mgh=20x9.8x10=1960J
Hence it will be same for both children because m,g and h are same.
But power=Work done/Time
For A, P=1960/10=196W
For B, P=1960/20=98W
Hence A has more power than B.

Q. At what position, pendulum acquires the maximum kinetic energy?

Ans: Mean position

When the pendulum reaches the very bottom of the swing i.e. the equilibrium/mean position, it is at its lowest point and greatest speed. This means that the pendulum has zero potential energy (with respect to its rest position) and maximum kinetic energy.

Q. Explain the transformation of energy in hydroelectric power plant?

Ans: In a hydroelectric power plant, the potential energy of water at higher level is converted to kinetic energy of the water by letting it flow at a faster rate. The fast falling water rotates the turbine of the generator. Thus, kinetic energy of the water is converted into mechanical energy of the turbine.
The axle of the turbine is connected to the armature of the electric generator. Thus the rotating turbine now rotates the armature in the magnetic field of the generator. As a result, mechanical energy of the rotating system of the generator is converted into electrical energy of the generator. Thus, we have electrical energy as output from the hydroelectric power plant.

Q. A boy weighing 50 kg climbs up a vertical height of 100m in 200 sec. Calculate the -
(a) amount of work done by him (b) potential energy gained by him
(c) power of the boy (g = 10m/s2)


Ans:
Mass of the boy, m = 50 kg Gravitational force on him = his weight = 50 × 10 = 500 N

(a) He has to apply a force equal to his weight to move upward.
                             So, work done = (500)(100) = 50000 J

(b)The work done by him on himself in going upward is stored in his body as potential energy.
So, PE gained by him = 50000 J

(c) We know, Power = work done/time taken = 50000/200 = 250 W

Related post: Numerical on work and energy for class IX                Read/View

Comments

Post a Comment

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca