# Frank Solutions for Chapter 1.4 Work, Energy, Power and their Relation with Force Class 10 Physics ICSE

**Exercise**

**1. Work is the application of a ....... through a distance.**

**Answer**

Force

**2. A boy does work when he pushes against a brick wall. (yes/no).**

**Answer**

Yes

**3. What is the SI unit of work?**

**Answer**

Joule

**4. Nm is the unit of .......**

**Answer**

Work

**5. One joule is the amount of work done when a force of _____ moves a body through a distance of ______**

**Answer**

1N, 1m in its own direction

**6. What is the work done when no net force is applied on the body?**

**Answer**

When no net force is applied, the work done which is the dot product of force and displacement is zero.

**7. What is the work done when the object acted upon by a force remains at rest?**

**Answer**

The work done is zero because the displacement is zero.

**8. What is the work done when the force on the object and the displacement of the object are perpendicular to each other?**

**Answer**

The work done is zero because :

work done = force × displacement × cosÎ¸

work done = force × displacement × cosÎ¸

**9. Is work done a scalar or a vector physical quantity?**

**Answer**

Work is a scalar quantity because it is a measure of transfer of energy without indicating any direction.

**10. What is the work done when the displacement of the body is in the opposite direction to that of the applied force?**

**Answer**

When the displacement of the body is in the direction opposite to that of the applied force, the work done is negative.

w = Fd cos Î¸ = Fd cos 180°

w = Fd cos Î¸ = Fd cos 180°

**11. What is the work done by the gravitational force of the earth on a satellite revolving**

**around the earth?**

**Answer**

The work done by the gravitational force of the earth on a satellite revolving around the earth is zero because the motion of the satellite is perpendicular to the force at every point.

**12. What is the work when a load of 500 kg is lifted vertically by 5 m? Given, g = 10 ms**

^{-2}. Express your answer in kilo joule.**Answer**

Given, load m = 500kg; height h = 5m; g = 10ms

W = mgh = 500×10×5 = 25000 J = 25 kJ

^{2}.W = mgh = 500×10×5 = 25000 J = 25 kJ

**13. A man lifts a mass of 10 kg from the floor to a shelf 4 meter high. If g = 10 ms**

^{-2}, what is the work done?**Answer**

Given, mass m = 10kg; height h = 4m; g = 10ms

W = mgh = 10×10×4 = 400J

^{2}.W = mgh = 10×10×4 = 400J

**14. What is the work done against gravity when a body is moved horizontally along a frictionless surface?**

**Answer**

The work done against gravity is zero when a body is moved horizontally along a frictionless surface because the force of gravity is perpendicular to the displacement in this case.

**15. What do you mean by the term 'work'?**

**Answer**

'Work' is said to be done when the applied force makes the body move i.e., there is a displacement of body.

It is equal to the product of force and the displacement of the point of application of the force in the direction of force.

It is equal to the product of force and the displacement of the point of application of the force in the direction of force.

**16. What are the quantities on which the work done depends?**

**Answer**

Work done depends upon:

(i) the magnitude and direction of the applied force, and

(ii) the displacement it produces.

(i) the magnitude and direction of the applied force, and

(ii) the displacement it produces.

**17. When you move upstairs, do you perform some work?**

**Answer**

Yes, we perform work against gravity.

**18. What should be the angle between the directions of the displacement and the applied force so that the work done is zero?**

**Answer**

The angle should be 90

^{o}.**19. Why is the work done on an object moving along a circular path zero?**

**Answer**

This is because at each point of the circular path, the displacement is perpendicular to the force, which is directed towards the centre, along the radius.

**20. In which one of the following cases is the work done more? When the angle between the direction of motion and that of the force is (i) 90° (ii) 0°.**

**Answer**

When the angle between the direction of motion and that of the force is 90°;

W = Fd cos 90° = 0

When the angle between the direction of motion and that of the force is 0°;

W = Fd cos 0° = Fd

Hence in the second case, when the angle is 0°; the work done is more.

**21. A man carrying a suitcase in his hand is walking horizontally. What is the work done against gravity?**

**Answer**

The displacement of the man and suitcase is along the horizontal direction. Thus, the angle between the displacement and the force of gravity is 90°;

Thus, W = Fd cos 90° = 0

Hence, no work is done against gravity in this case.

**22. What is the work done on a body by the gravitational force towards the centre of the path acting when it moves along a circular path?**

**Answer**

When a body moves along a circular path, work done by the gravitational force towards the centre of the path is zero, because the displacement in this case is normal to the gravitational force.

**23. What is the work done on the earth by the gravitational force of the sun during its motion around the sun?**

**Answer**

The work done by the gravitational force of the sun on earth during its motion around the sun is zero because at every point, the displacement of earth is perpendicular to the gravitational force of sun i.e.,

W = Fd cos 90

W = Fd cos 90

^{0}= 0**24. What do you mean by a kilo joule?**

**Answer**

A kilojoule of work is said to done when a force of 1 newton displaces a body through 1000 metres in its own direction.

1 kJ = 10

1 kJ = 10

^{3}joules**25. How many joules are there in 1 mega joule?**

**Answer**

1 MJ = 10

^{6}joules**26. Define 'joule'.**

**Answer**

The SI unit of work is joule.

1 joule of work is said to be done when a force of 1 newton displaces a body through 1 metre in its own direction.

1 joule of work is said to be done when a force of 1 newton displaces a body through 1 metre in its own direction.

**27. What is the ratio of SI units to CGS units of 'work'?**

**Answer**

The SI unit of work is 'joules' and the CGS unit is 'erg'.

1 joule = 10

Thus the ratio is 10

1 joule = 10

^{7}ergThus the ratio is 10

^{7}: 1**28. An engine does 54,000 J of work by exerting a force of 6000 N on it. What is the displacement in the direction of the force?**

**Answer**

Given, work done = 54000 J, force = 6000N, Î¸ = 0°

Now, work done = force × displacement × cos0°

or, displacement = 54000/6000 = 9m

Now, work done = force × displacement × cos0°

or, displacement = 54000/6000 = 9m

**29. How much force is applied on a body when 150 J of work is done in displacing the body through a distance of 10 m in the direction of the force?**

**Answer**

Given, work done = 150 J, displacement = 10m, Î¸ = 0°

Now, work done = force × displacement × cos0°

or, force = 150/10 =15 N

Now, work done = force × displacement × cos0°

or, force = 150/10 =15 N

**30. Work done by a force is equal to the product of ____ and ______**

**Answer**

Work done by a force is equal to the product of

__applied force__and__displacement in the direction of the applied force__.**31. Give two examples of work done.**

**Answer**

Examples of work done:

- In free fall of a body of mass m, under gravity from a height h, the force of gravity (F=mg) is in the direction of displacement (h) and the work done by the gravity is mgh.
- A coolie lifting a load does work against gravity.

**32. On what factors does the work done by a force depend?**

**Answer**

Work done depends upon:

- the magnitude and direction of the applied force, and
- the displacement it produces.

**33. Write an expression for work done by a force depends?**

**Answer**

Work done = force × displacement × cos Î¸

or, W = Fd cos Î¸

or, W = Fd cos Î¸

**34. Write an expression for the work done against gravity.**

**Answer**

Work done against gravity = mass × acceleration due to gravity ×height

Or, W = mgh

Or, W = mgh

**35. No work is done by a man moving on a horizontal road while carrying a box on his head. Explain.**

**Answer**

The displacement of the man and box is along the horizontal direction. Thus, the angle between the displacement and the force of gravity is 90°;

Thus, W = Fd cos 90° = 0

Thus, W = Fd cos 90° = 0

Hence, no work is done against gravity in this case; however some work is done against friction.

**36. Is power a scalar quantity?**

**Answer**

Yes, power is a scalar quantity.

**37. Can every force produce work?**

**Answer**

No, every force cannot produce work. Force can produce work if the applied force cause displacement in the direction of the force.

**38. Distinguish between work and power.**

**Answer**

Work is said to be done only when the applied force on a body makes the body move but power is the rate of doing work.

The SI unit of work is 'joules' and that of power is 'watt'.

The SI unit of work is 'joules' and that of power is 'watt'.

**39. Complete the following sentences:**

(a) The SI unit of work is _____ and of power is _____

(b) Kilowatt is the unit of _____ and kWh is the unit of

(c) Joule is the unit of _____

(d) 1 J = _____ Erg.

(e) 1 H.P. = _____ W.

(a) The SI unit of work is _____ and of power is _____

(b) Kilowatt is the unit of _____ and kWh is the unit of

(c) Joule is the unit of _____

(d) 1 J = _____ Erg.

(e) 1 H.P. = _____ W.

**Answer**

(a) joule, watt

(b) power, energy

(c) work

(d) 10

^{7}(e) 746

**40. A weight lifted a load of 200 kgf to a height of 2.5 m in 5 s. calculate:**

(i) the work done, and

(ii) the power developed by him. Take g= 10 N kg

(i) the work done, and

(ii) the power developed by him. Take g= 10 N kg

^{-1}.**Answer**

Given, load m = 200 kgf, displacement h = 25 m , time = 5s, g = 10 Nkg

Now, work done = 200 kgf × 25 = 5000 J

or , power = work done/time taken = 5000/5 = 1000 watt

^{-1}.Now, work done = 200 kgf × 25 = 5000 J

or , power = work done/time taken = 5000/5 = 1000 watt

**43. A boy of mass m climbs up a staircase of vertical height h. What is the work done by the boy against the force of gravity? What would have been the work done if he uses a lift in climbing the same vertical height?**

**Answer**

Work done depends upon the vertical height and not the path taken, hence if the boy uses a lift to reach the same vertical height, work done will be mgh.

**44. Can work done be zero even if force acts on the body?**

**Answer**

Yes, for e.g. if you push a wall, you apply force on it but no work is done since the displacement is zero.

**45. What is the relation between 1 H.P. and 1 kW?**

**Answer**

1 H.P. = 0.746 kW

**46. It takes 20 s for A to climb up the stairs, while B does the same in 15 s. compare the**

(i) work done, and

(ii) power developed by A and B.

(i) work done, and

(ii) power developed by A and B.

**Answer**

(i) Work done depends on the displacement, now, vertical distance travelled by A and B is same, hence ratio of work done by them is 1 : 1

(ii) Power = work done/time taken

(ii) Power = work done/time taken

**47. State and define the SI unit of power.**

**Answer**

The SI unit of power is 'watt'.

If 1 joule of work is done in 1 second, the power is said to be 1 watt.

If 1 joule of work is done in 1 second, the power is said to be 1 watt.

**48. Write the SI and CGS units of power. How are they related?**

**Answer**

SI unit of power = watt

CGS unit of power = erg per second

1W = 1 Js

CGS unit of power = erg per second

1W = 1 Js

^{-1 }= 10^{7}erg s^{-1}.**41. A machine raises a load of 750 N through a height of 16 m in 5s. calculate:**

(i) work done by machine,

(ii) power at which the machine works.

(i) work done by machine,

(ii) power at which the machine works.

**Answer**

Given, load M = 750 N, displacement h = 16m, time 5s, g = 10 N kg

^{-1}.Now, work done = mgh = 750 × 16 = 12000 J

or, power = work done/time taken = 12000/5 = 2400 watt

**42. Name the physical quantity whose MKS units are kgm**

^{2}s^{-3}.**Answer**

Power