Work and Power Numerical Class 11 Nootan ISC Physics Solutions Ch-9. Step by step Solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Work and Power Numerical Class 11 ISC Nootan Solutions Ch-9 of Kumar and Mittal Physics , Nageen Prakashan
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Writer | Kumar and Mittal |
| Publication | Nageen Prakashan |
| Chapter-9 | Work , Energy and Power |
| Topics | Numericals on Work and Power |
| Academic Session | 2024-2025 |
Numericals on Work and Power
Ch-9 Numerical Class 11 ISC Nootan Solutions, Kumar and Mittal Physics , Nageen Prakashan
Question-1: A 55 kg man holds a weight of 20 kg on his head. What is the work done by him against gravity if he moves a distance of 20 m (i) on a horizontal road, (ii) on an incline of 1 in 5? Take g = 10 m s^-2.
Answer- (i) in first case work done is zero as displacement (horizontal) is at right angle from force (weight) vertical
(ii) let he acquires height h then
=> h/20 = 1/5
=> h = 4
again work done = gain energy = m g h
=> (55 + 20) x 10 x 4 = 3000 J.
Question-2: find the work done when a 25 kg weight is (i) lifted to vertical height of 2.0 m from the ground, (ii) carried to the same place by pushing it up an inclined plane making an angle of 30° with the ground. Take g = 9.8 m s^-2.
Answer- (i) W =m g h
=> 25 x 9.8 x 2 = 490 J
(ii) since gravitational force is a conservative force
∴ work done in both the situation will be same i.e. 490 J.
Question-3: A 10 kg weight is raised to a height of 2.0 m with an acceleration of 1.5 m s^-2. Compute the work done, if g=10 m s^-2.
Answer- Net upward acceleration
=> a + g = 1.5 + 10 = 11.5 m/s²
∴ Force = m x a = 10 x 11.5 = 115 N
∴ work done = 115 x 2 = 230 J
Question-4: A 25 kg box is pulled at a constant speed by a horizontal force on a horizontal floor. The coefficient of sliding friction between the box and the floor is 0.20. How much work is done in pulling the box through 20 m? (g = 9.8 m s^-2)
Answer- F = μmg = 0.2 x 25 x 9.8 = 49 N
again work done = F x s
=> 49 x 20 = 980 J
Question-5 Find the power of a 60 kg man who can climb up a height of 10 m in half a minute. g = 9.8 m s^2.
Answer- P = W/t
=> 60 x 9.8 x 10 /30 = 196 Watt.
Question-6: The power of a pump-motor is 2 kW. How much water per minute can it raise to a height of 10 metre? (g = 10 m/s²)
Answer- P =W/t
=> W = P x t
=> m g h = P x t
=> m = P x t / g h
=> 2000 x 100 / 10 x 10 = 1200 kg
Question-7: A man pulls a roller by a force of 20 kg-f applied at 60° with the ground. If he pulls it a distance of 10 m in 1 minute, calculate the power dissipated. Take g = 10 m s^2
Answer- P = W/t
=> (20 x 10 x 10 x 1/2) /60 = 16.7 Watt
Question-8: An engine can pull 500 metric ton load up an inclined plane rising 1 in 100 with a speed of 10 m/s. The frictional force offered by the plane is 2000 N. What is the power of the engine?
Answer- Net force = component of weight of engine along plain + friction
i.e. 500 x 10³ g sinθ + 2000
=> 500 x 10³ x 9..8 x 1/100 + 2000
=> 49000 + 2000 = 51000 N
again power = F x v
=> 51000 x 10
=> 510 x 1000 W or 510 kW.
Question-9: An engine pulls a 1500 kg car on a level road at a constant speed of 5.0 m/s against a frictional force of 500 N. Calculate the power expended by the engine. What extra power has the engine to expend in order to maintain the same speed of the car up an inclined plane having a gradient of 1 in 10?
Answer- m = 1500 kg , v = 5 m/s , F = 500N
=> P = F x v
=> 500 x 5 = 2500 W = 2.5 kW
again increase of inclined plane components of weight of car along plain will be added to force
=> F = 500 + 1500 x 9.8 x 1/10 = 1970 N
∴ New Power = 1970 x 5 = 9850 W = 9.85 kW
∴ extra power = 9.85 – 2.5 = 7.35 kW
—: end of Work and Power Numerical Class 11 ISC Nootan Solutions Ch-9 Kumar and Mittal Physics :—
Return to : – Nootan Solutions for ISC Physics Class-11 Nageen Prakashan
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