Extra Questions for Class 9th: Ch 11 Work and Energy (Science) Important Questions Answer Included
Very Short Answer Questions (VSAQs): 1 MarkQ1. What is the work done against the gravity when a body is moved horizontally along a frictionless surface?
Answer
Zero
Q2. When displacement is in a direction opposite to the direction of force applied, what is the type of work done?
Answer
Q2. When displacement is in a direction opposite to the direction of force applied, what is the type of work done?
Answer
Negative work.
Q3. A 40 kg girl is running along a circular path of radius 1 m with a uniform speed. How much work is done by the girl in completing one circle?
Answer
Q3. A 40 kg girl is running along a circular path of radius 1 m with a uniform speed. How much work is done by the girl in completing one circle?
Answer
Zero
Q4. Seema tried to push a heavy rock of 100 kg for 200 s but could not move it. Find the work done by Seema at the end of 200 s.
Answer
Q4. Seema tried to push a heavy rock of 100 kg for 200 s but could not move it. Find the work done by Seema at the end of 200 s.
Answer
Work done = 0 Since displacement, s = 0
Q5. Identify the kind of energy possessed by a running athlete.
Answer
Q5. Identify the kind of energy possessed by a running athlete.
Answer
Kinetic energy.
Answer
Answer
Answer
Short Answer Questions-II (SAQs-II) : 3 Marks
Q1. When a force retards the motion of a body, what is the nature of work done by the force? State reason. List two examples of such a situation.
Answer
𝐹=8𝑁
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒, 𝑊=𝐹×𝑠
𝑊1= 8×20=160 𝐽
𝐷=10 𝑚
S𝑜 𝑟𝑎𝑑𝑖𝑢𝑠, 𝐷/2=5𝑚
𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒=2 πr
=2×22/7×5
=31.43
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 1⁄2 𝑐𝑖𝑟𝑐𝑙𝑒= πr
= 22/7×5=15.71
𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑜𝑟 11⁄2 𝑐𝑖𝑟𝑐𝑙𝑒=31.43+15.71=47.14 𝑚
𝑊2=𝐹×𝑠=8 ×47.14=376 𝐽
𝑊3=20×8=160 𝐽
𝑇𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒=160+376+160
696 𝐽
Long Answer Questions (LAQs) : 5 Marks
Q1. Calculate the electricity bill amount for a month of 30 days, if the following devices are used as specified:
(i) 2 bulbs of 40 W for 6 hours.
(ii) 2 tubelights of 50 W for 8 hours.
(iii) A TV of 120 W for 6 hours.
Given the cost of electricity is ₹2.50 per unit.
Answer
Short Answer Questions-I (SAQs-I) : 2 Marks
Q1. An electrical heater is rated 1200 W. How much energy does it use in 10 hours?
Q1. An electrical heater is rated 1200 W. How much energy does it use in 10 hours?
Answer
Electrical energy = Power × time taken
= 1.2 × 10 = 12 kWh
Q2. If an electric appliance is rated 1000 W and is used for 2 hours. Calculate the work done in 2 hours.
= 1.2 × 10 = 12 kWh
Q2. If an electric appliance is rated 1000 W and is used for 2 hours. Calculate the work done in 2 hours.
Answer
Work done = Energy consumed
Energy = Power × Time taken
= 1000 W × 2 hour
= 2000 W-hr or 2 kW-hour or 2 kWh
Q3. A man of mass 62 kg sums up a stair case of 65 steps in 12 s. If height of each step is 20 cm, find his power.
Answer
Energy = Power × Time taken
= 1000 W × 2 hour
= 2000 W-hr or 2 kW-hour or 2 kWh
Q3. A man of mass 62 kg sums up a stair case of 65 steps in 12 s. If height of each step is 20 cm, find his power.
Answer
𝑃.𝐸. =𝑚𝑔ℎ
𝑚=62 𝑘𝑔, 𝑔=10 𝑚/𝑠2 , ℎ=65 × 20/100 =13 𝑚
𝑃.𝐸. =62 ×10 ×13 = 8060 𝐽
𝑃= (𝑃.𝐸.)/𝑡= 8060/12
= 671.67 𝑊
Q4. How is work done by a force measured? A porter lifts a luggage of 20 kg from the ground and puts it on his head 1.7 m above the ground. Find the work done by the porter on the luggage. (𝒈=𝟏𝟎𝒎/𝒔2)
Answer
𝑚=62 𝑘𝑔, 𝑔=10 𝑚/𝑠2 , ℎ=65 × 20/100 =13 𝑚
𝑃.𝐸. =62 ×10 ×13 = 8060 𝐽
𝑃= (𝑃.𝐸.)/𝑡= 8060/12
= 671.67 𝑊
Q4. How is work done by a force measured? A porter lifts a luggage of 20 kg from the ground and puts it on his head 1.7 m above the ground. Find the work done by the porter on the luggage. (𝒈=𝟏𝟎𝒎/𝒔2)
Answer
Work done is product of force and displacement .
𝑊=𝐹 ×𝑠
𝑚=20 𝑘𝑔,
𝑊=𝐹 ×𝑠
𝑚=20 𝑘𝑔,
𝑔=10𝑚/𝑠2,
ℎ =1.7𝑚
𝑇ℎ𝑒 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑝𝑜𝑟𝑡𝑒𝑟=𝑚𝑔ℎ
=20×10×1.7=340 𝐽
Q5. The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.
𝑇ℎ𝑒 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑝𝑜𝑟𝑡𝑒𝑟=𝑚𝑔ℎ
=20×10×1.7=340 𝐽
Q5. The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.
Answer
𝑣2−𝑢2=2𝑎𝑠
𝑠=[(𝑣2−𝑢2)/2𝑎]
𝑠=[(𝑣2−𝑢2)/2𝑎]
𝐹=𝑚𝑎
𝑊=𝐹×𝑠
𝑊=𝑚𝑎 [(𝑣2−𝑢2)/2𝑎]
𝑊=𝐹×𝑠
𝑊=𝑚𝑎 [(𝑣2−𝑢2)/2𝑎]
= 1/2 𝑚𝑣2 −1/2 𝑚𝑢2
=(𝐾.𝐸.)final − (𝐾.𝐸.)initial
=(𝐾.𝐸.)final − (𝐾.𝐸.)initial
Short Answer Questions-II (SAQs-II) : 3 Marks
Q1. When a force retards the motion of a body, what is the nature of work done by the force? State reason. List two examples of such a situation.
Answer
The work done by the force is negative because the displacement is opposite to the direction of force applied.
Example: (i) Work done by the force of friction;
(ii) Work done by applying brakes.
Q2. When is the work done by a force said to be negative ? Give one situation in which one of the forces acting on the object is doing positive work and the other is doing negative work.
Answer
Example: (i) Work done by the force of friction;
(ii) Work done by applying brakes.
Q2. When is the work done by a force said to be negative ? Give one situation in which one of the forces acting on the object is doing positive work and the other is doing negative work.
Answer
Negative work: When the force is acting opposite to the direction of the displacement, the work done by the force is said to be negative. When we lift an object, two forces act on the
(i) Muscular force: Doing positive work in the direction of the displacement.
(ii) Gravitational force: Doing negative work opposite to the direction of the displacement.
Q3. (a) Under what conditions work is said to be done?
(b) A porter lifts a luggage of 1.5 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer
(i) Muscular force: Doing positive work in the direction of the displacement.
(ii) Gravitational force: Doing negative work opposite to the direction of the displacement.
Q3. (a) Under what conditions work is said to be done?
(b) A porter lifts a luggage of 1.5 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer
(a) (i) Force should be applied.
(ii) Body should move in the line of action of force.
(iii)Angle between force and displacement should not be 90°.
(b) Mass of luggage, m = 15 kg and displacement, s = 1.5 m.
(ii) Body should move in the line of action of force.
(iii)Angle between force and displacement should not be 90°.
(b) Mass of luggage, m = 15 kg and displacement, s = 1.5 m.
Work done, W = F×s = mg × s = 15 × 10 × 1.5 = 225 J
Q4. Four persons jointly lift a 250 kg box to a height of 1 m and hold it.
Q4. Four persons jointly lift a 250 kg box to a height of 1 m and hold it.
(i) Calculate the work done by the persons in lifting the box.
(ii) How much work is done for just holding the box ?
(iii) Why do they get tired while holding it ? (𝑔=10𝑚𝑠2)
Answer
(ii) How much work is done for just holding the box ?
(iii) Why do they get tired while holding it ? (𝑔=10𝑚𝑠2)
Answer
(i) 𝐹=250×10=2500𝑁
𝑠=1 𝑚
𝑊=𝐹 ×𝑠=2500×1=2500𝐽
(ii) Zero, as there is no displacement.
(iii) To hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. While applying the force muscular effort is involved, and so they feel tired.
Q5. A boy is pulling a cart by supplying a constant force of 8 N on a straight path of 20 m. On a round about of 10 m diameter he forgets the path and takes 1½ turns and then continues on the straight path for another 20 m. Find the net work done by the boy on the cart.
𝑠=1 𝑚
𝑊=𝐹 ×𝑠=2500×1=2500𝐽
(ii) Zero, as there is no displacement.
(iii) To hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. While applying the force muscular effort is involved, and so they feel tired.
Q5. A boy is pulling a cart by supplying a constant force of 8 N on a straight path of 20 m. On a round about of 10 m diameter he forgets the path and takes 1½ turns and then continues on the straight path for another 20 m. Find the net work done by the boy on the cart.
Answer
𝐹=8𝑁
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒, 𝑊=𝐹×𝑠
𝑊1= 8×20=160 𝐽
𝐷=10 𝑚
S𝑜 𝑟𝑎𝑑𝑖𝑢𝑠, 𝐷/2=5𝑚
𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒=2 πr
=2×22/7×5
=31.43
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 1⁄2 𝑐𝑖𝑟𝑐𝑙𝑒= πr
= 22/7×5=15.71
𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑜𝑟 11⁄2 𝑐𝑖𝑟𝑐𝑙𝑒=31.43+15.71=47.14 𝑚
𝑊2=𝐹×𝑠=8 ×47.14=376 𝐽
𝑊3=20×8=160 𝐽
𝑇𝑜𝑡𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒=160+376+160
696 𝐽
Long Answer Questions (LAQs) : 5 Marks
Q1. Calculate the electricity bill amount for a month of 30 days, if the following devices are used as specified:
(i) 2 bulbs of 40 W for 6 hours.
(ii) 2 tubelights of 50 W for 8 hours.
(iii) A TV of 120 W for 6 hours.
Given the cost of electricity is ₹2.50 per unit.
Answer
Given the cost of electricity is ₹2.50 per unit.
(i) 2 bulbs of 40 watts for 6 hrs.
Energy consumed by Bulbs 𝐸1 =2 ×40×6=480 𝑊=0.48 𝐾𝑤ℎ
(ii) Energy consumed by 2 tubelights 𝐸2=50×8 ×2=0.800 𝑘𝑊ℎ
(iii) 𝐸nergy consumed by TV E3=120×6=0.720 𝑘𝑊ℎ
𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 0.48+0.80+0.72=2.00 𝑢𝑛𝑖𝑡𝑠 𝑟𝑎𝑡𝑒=2.50 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝑑𝑎𝑦 = 2 ×2.50=5.00
𝐶𝑜𝑠𝑡 30 𝑑𝑎𝑦𝑠 = 5.00×30=150
Q2. (i) What is meant by mechanical energy? State its two forms. State the law of conservation of energy. Give an example in which we observe a continuous change of one form of energy into another and vice-versa.
(ii) Calculate the amount of work required to stop a car of 1000 kg moving with a speed of 72 km/h.
Answer
Energy consumed by Bulbs 𝐸1 =2 ×40×6=480 𝑊=0.48 𝐾𝑤ℎ
(ii) Energy consumed by 2 tubelights 𝐸2=50×8 ×2=0.800 𝑘𝑊ℎ
(iii) 𝐸nergy consumed by TV E3=120×6=0.720 𝑘𝑊ℎ
𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 0.48+0.80+0.72=2.00 𝑢𝑛𝑖𝑡𝑠 𝑟𝑎𝑡𝑒=2.50 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝑑𝑎𝑦 = 2 ×2.50=5.00
𝐶𝑜𝑠𝑡 30 𝑑𝑎𝑦𝑠 = 5.00×30=150
Q2. (i) What is meant by mechanical energy? State its two forms. State the law of conservation of energy. Give an example in which we observe a continuous change of one form of energy into another and vice-versa.
(ii) Calculate the amount of work required to stop a car of 1000 kg moving with a speed of 72 km/h.
Answer
(i) Sum of kinetic energy and potential energy of an object is the total mechanical energy.
Its two forms are kinetic energy and potential energy.
Energy can neither be created nor be destroyed but can be transformed from one form to another. 1 Example is simple pendulum.
(ii) 𝑚=1000 𝑘𝑔,
Its two forms are kinetic energy and potential energy.
Energy can neither be created nor be destroyed but can be transformed from one form to another. 1 Example is simple pendulum.
(ii) 𝑚=1000 𝑘𝑔,
𝑢=72 𝑘𝑚/ℎ =72× 5/18 𝑚/𝑠 =20 𝑚/𝑠,𝑣=0,
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 = 1/2 𝑚𝑣2
= 1/2 ×1000×202
= 200000 𝐽 = 2 × 105 𝐽
𝐹𝑖𝑛𝑎𝑙 𝐾.𝐸. = 0
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 = 1/2 𝑚𝑣2
= 1/2 ×1000×202
= 200000 𝐽 = 2 × 105 𝐽
𝐹𝑖𝑛𝑎𝑙 𝐾.𝐸. = 0
W𝑜𝑟𝑘 𝑑𝑜𝑛𝑒=𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐾.𝐸. =2×105 𝐽