Positive Work : From the mathematical expression of work
When θ is acute angle i.e., less than 90°, then Cos θ is +ve
i.e., if θ < 90, Cos θ = +ve and work done will also be positive
i.e., W = F.S Cos θ = +ve
Work done is said to be +ve if the angle θ between applied force F and the displacement S is a acute angle i.e., θ < 90°.
Examples of positive work done :
(i) When a body falls freely under gravitational pull, the work done by gravity is positive.
(ii) When a horse pulls a cart on a level road, the work done is positive.
(iii) When a string is stretched, the stretching force acts in the direction of displacement, here work done by stretching force is positive.
In all the above examples F and S are in the same direction hence work done is +ve.
Negative Work : From the mathematical expression of work
W = F.S Cos θ
When θ > 90°, Cos θ is –ve, hence work is also negative. Thus work done is said to be negative if angle(θ) between applied force (F) and the displacement (S) is an obtuse angle i.e., θ > 90°.
Examples of negative work done :
(i) When a body is made to slide over a surface, the work done by the frictional forces is negative.
(ii) When two similar poles of a magnet are brought close to each other, the work done by the magnetic force of repulsion is negative.
(iii) When two similar charges either both positive or both negative are brought close to each other, they repel each other hence the work done by electrostatic force of repulsion acting between them will be negative.
In all the above examples F and S are in the opposite directions, hence work done is –ve.
Zero Work : From the mathematical expression of work
W = F.S Cos θ
Work done by a force will be zero when either θ = 90° or S = 0 i.e., if there is no displacement.
Hence, work done will be zero if either angle (θ) between applied force (F) and the displacement (S) is 90° or the body does not move under the application of force.
Examples of zero work done :
(i) When a person travels on a platform with a load on his head, work done by him is zero.
(ii) when a body moves along a circular path with the help of a string, the work done by tension in the string is zero.
(iii) When a man tending to push a wall, he is doing no work. though the force is being applied, because the point to application of force i.e., wall does not move.
In the above example, displacement is zero hence work done is also zero, while in the example (i) and (ii), angle between F and S. i.e., θ = 90°. Hence work is zero.
Practice Problems:
1. What is the work done by a man in carrying a suitcase weighing 30 kg over his head when he travels a distance of 10 m in :
(i) the direction direction;
(ii) vertically upward direction.
Sol. (i) Work done = F Cos θ × S
Here, F = force exerted by the man = mg = 30 × 9.8 N
θ = 90°, Since the body moves in horizontal direction and force acts vertically upwards.
∴ Cos θ = Cos 90° = 0;
S = 10 m
∴ W.D = 30 × 9.8 × 0 × 10
= 0
( No work is done when the suitcase of 30 kg is moved by a man on his head in the horizontal direction)
(ii) Work done in moving in vertical upwards direction = mgh
Here, m = 30 Kg;
g = 9.8 m/s2
∴ W.D = 30 × 9.8 × 10 J
= 2940 J (Ans.)
2940 J work is done by a man in carrying a suitcase of 30 Kg on his head in the vertical upward direction.
2. A man weighing 50 Kg caries a load of 10 kg on his head. Find the W.D. when he goes (i) 15 m vertically up
(iii) 15 m on a leveled path on the ground.
Sol. Here, Mass of the man, m1 = 50 Kg
Mass carried by the man , m2 = 10 kg
Total mass, M = m1 + m2
= 50 + 10
= 60 kg.
(i) When the man goes vertically up,
Height through which rises, h = 15 m.
∴ W.D. = Mgh = 60 × 9.8 × 15
= 8820 J (Ans.)
(ii) When the man goes on a leveled path on the ground.
W.D. = F Cos θ × S
Here, θ = 90°
∴ Cos θ = Cos 90° = 0;
S = 15 m. ∴ W.D, = F × 0 × 15 = 0
(No work done in moving on a leveled path i.e., horizontally)
3. A body moves through a horizontal distance of 20 m under the action of a force of 1.5 N. Calculate the work done if the force makes an angle of
(i) 30° with the horizontal
(ii) 60° with the horizontal.
Sol. (i) Here,
Force, F = 15 N and distance, S = 20 m
Angle with horizontal, θ = 30°
∴ work done is given by;
W = F Cos θ × S
Therefore, work done is given by,
W = F Cos θ × S
= 15 N × cos 60° × 20 m
= 15 N × 1/2 × 20 m
= 150 J (Ans.)
4. Calculate the amount of work done when 1 kg water falls through the height 10 m.
Sol. Work done = mgh = 1× 9.8× 10
= 98 J (Ans.)
5. A body of mass 20 kg initially at rest is moved by a horizontal force of 5N on a smooth frictionless table. What will be work done by the force in 8 sec. ?
Sol. Given, u = 0 , t = 8
a = F/m = 5/20
We know that, S= ut + (1/2) at2
= 0 × 8 + (1/2) × (5/20) × (8)2
S = 1/2 × 5/20 × 64
= 8 m
Work done = FS = 5 × 8 = 40 J. (Ans.)
Check your understanding with Multiple choice questions:
1. Work is done when force
(a) is applied
(b) produces motion
(c) is normal to displacement
(d) is strong
2. When the angle between force and displacement is obtuse, the work done is
(a) 1 (b) Zero
(c) +ve (d) –ve
3. When we lower a bucket into a well to fetch the water. The work done is
(a) 1 (b) 0
(c) +ve (d)–ve
4. Work is measured in
(a) Newton (b) Joule
(c) Watt (d) Newton-cm
5. A stone of mass 0.1 Kg is whirled in a horizontal circle of radius 21cm with a force of 100 dyne work done to complete one cycle is
(a) 0 (b) 100 J
(c) 100 dyne-cm (d) 220 dyne-cm
6. The sun's gravitational force is keeping the earth in its orbit. The work done by this force is
(a) 0 (b) +ve
(c) –ve (d) Infinite
7. Which of the following is not related with work ?
(a) F.S (b)Nm
(c) Joule (d)Hz.
8. Which of the following is not a unit of work?
(a) Joule (b) Kg m
(c) g cm (d) All are units of work.
9. Choose the incorrect statement .........
(a) Work done by centripetal force is Zero
(b) When a body slides on a horizontal surface, no work is done either by weight (mg) or normal reaction (R)
(c) Work done by frictional force is always negative
(d) A person walking on a horizontal road with a load on his head is doing work.
10. Work is
(a) Scalar quantity
(b) Vector quantity
(c) Neutral quantity
(d) F × S
11. A man pushes a wall but fails to displace it; he does
(a) negative work
(b) positive work
(c) maximum work
(d) No work at all
12. The work performed on an object does not depend on-
(a) applied Force
(b) displacement
(c) angle at which force is inclined to displacement
(d) initial velocity of the object.
13. A body carrying a box on his head is walking on level road from one place to another on a straight road is doing no work. This statement is -
(a) correct
(b) incorrect
(c) partly correct
(d) insufficient data
Answers:
1. b
2. d
3. d
4. b
5. a
6. a
7. d
8. d
9. d
10. a
11. d
12. d
13. a
W = F.S Cos θ
Where θ is the smallest angle between F and the displacement S.When θ is acute angle i.e., less than 90°, then Cos θ is +ve
i.e., if θ < 90, Cos θ = +ve and work done will also be positive
i.e., W = F.S Cos θ = +ve
Work done is said to be +ve if the angle θ between applied force F and the displacement S is a acute angle i.e., θ < 90°.
Examples of positive work done :
(i) When a body falls freely under gravitational pull, the work done by gravity is positive.
(ii) When a horse pulls a cart on a level road, the work done is positive.
(iii) When a string is stretched, the stretching force acts in the direction of displacement, here work done by stretching force is positive.
In all the above examples F and S are in the same direction hence work done is +ve.
Negative Work : From the mathematical expression of work
W = F.S Cos θ
When θ > 90°, Cos θ is –ve, hence work is also negative. Thus work done is said to be negative if angle(θ) between applied force (F) and the displacement (S) is an obtuse angle i.e., θ > 90°.
Examples of negative work done :
(i) When a body is made to slide over a surface, the work done by the frictional forces is negative.
(ii) When two similar poles of a magnet are brought close to each other, the work done by the magnetic force of repulsion is negative.
(iii) When two similar charges either both positive or both negative are brought close to each other, they repel each other hence the work done by electrostatic force of repulsion acting between them will be negative.
In all the above examples F and S are in the opposite directions, hence work done is –ve.
Zero Work : From the mathematical expression of work
W = F.S Cos θ
Work done by a force will be zero when either θ = 90° or S = 0 i.e., if there is no displacement.
Hence, work done will be zero if either angle (θ) between applied force (F) and the displacement (S) is 90° or the body does not move under the application of force.
Examples of zero work done :
(i) When a person travels on a platform with a load on his head, work done by him is zero.
(ii) when a body moves along a circular path with the help of a string, the work done by tension in the string is zero.
(iii) When a man tending to push a wall, he is doing no work. though the force is being applied, because the point to application of force i.e., wall does not move.
In the above example, displacement is zero hence work done is also zero, while in the example (i) and (ii), angle between F and S. i.e., θ = 90°. Hence work is zero.
Practice Problems:
1. What is the work done by a man in carrying a suitcase weighing 30 kg over his head when he travels a distance of 10 m in :
(i) the direction direction;
(ii) vertically upward direction.
Sol. (i) Work done = F Cos θ × S
Here, F = force exerted by the man = mg = 30 × 9.8 N
θ = 90°, Since the body moves in horizontal direction and force acts vertically upwards.
∴ Cos θ = Cos 90° = 0;
S = 10 m
∴ W.D = 30 × 9.8 × 0 × 10
= 0
( No work is done when the suitcase of 30 kg is moved by a man on his head in the horizontal direction)
(ii) Work done in moving in vertical upwards direction = mgh
Here, m = 30 Kg;
g = 9.8 m/s2
∴ W.D = 30 × 9.8 × 10 J
= 2940 J (Ans.)
2940 J work is done by a man in carrying a suitcase of 30 Kg on his head in the vertical upward direction.
2. A man weighing 50 Kg caries a load of 10 kg on his head. Find the W.D. when he goes (i) 15 m vertically up
(iii) 15 m on a leveled path on the ground.
Sol. Here, Mass of the man, m1 = 50 Kg
Mass carried by the man , m2 = 10 kg
Total mass, M = m1 + m2
= 50 + 10
= 60 kg.
(i) When the man goes vertically up,
Height through which rises, h = 15 m.
∴ W.D. = Mgh = 60 × 9.8 × 15
= 8820 J (Ans.)
(ii) When the man goes on a leveled path on the ground.
W.D. = F Cos θ × S
Here, θ = 90°
∴ Cos θ = Cos 90° = 0;
S = 15 m. ∴ W.D, = F × 0 × 15 = 0
(No work done in moving on a leveled path i.e., horizontally)
3. A body moves through a horizontal distance of 20 m under the action of a force of 1.5 N. Calculate the work done if the force makes an angle of
(i) 30° with the horizontal
(ii) 60° with the horizontal.
Sol. (i) Here,
Force, F = 15 N and distance, S = 20 m
Angle with horizontal, θ = 30°
∴ work done is given by;
W = F Cos θ × S
= 15 N × cos 30° × 20 m
= 300 × √3/2 Nm
= 259.8 joules (Ans.)
(ii) Here, F = 15 N, S = 20 m, θ = 60°Therefore, work done is given by,
W = F Cos θ × S
= 15 N × cos 60° × 20 m
= 15 N × 1/2 × 20 m
= 150 J (Ans.)
4. Calculate the amount of work done when 1 kg water falls through the height 10 m.
Sol. Work done = mgh = 1× 9.8× 10
= 98 J (Ans.)
5. A body of mass 20 kg initially at rest is moved by a horizontal force of 5N on a smooth frictionless table. What will be work done by the force in 8 sec. ?
Sol. Given, u = 0 , t = 8
a = F/m = 5/20
We know that, S= ut + (1/2) at2
= 0 × 8 + (1/2) × (5/20) × (8)2
S = 1/2 × 5/20 × 64
= 8 m
Work done = FS = 5 × 8 = 40 J. (Ans.)
Check your understanding with Multiple choice questions:
1. Work is done when force
(a) is applied
(b) produces motion
(c) is normal to displacement
(d) is strong
2. When the angle between force and displacement is obtuse, the work done is
(a) 1 (b) Zero
(c) +ve (d) –ve
3. When we lower a bucket into a well to fetch the water. The work done is
(a) 1 (b) 0
(c) +ve (d)–ve
4. Work is measured in
(a) Newton (b) Joule
(c) Watt (d) Newton-cm
5. A stone of mass 0.1 Kg is whirled in a horizontal circle of radius 21cm with a force of 100 dyne work done to complete one cycle is
(a) 0 (b) 100 J
(c) 100 dyne-cm (d) 220 dyne-cm
6. The sun's gravitational force is keeping the earth in its orbit. The work done by this force is
(a) 0 (b) +ve
(c) –ve (d) Infinite
7. Which of the following is not related with work ?
(a) F.S (b)Nm
(c) Joule (d)Hz.
8. Which of the following is not a unit of work?
(a) Joule (b) Kg m
(c) g cm (d) All are units of work.
9. Choose the incorrect statement .........
(a) Work done by centripetal force is Zero
(b) When a body slides on a horizontal surface, no work is done either by weight (mg) or normal reaction (R)
(c) Work done by frictional force is always negative
(d) A person walking on a horizontal road with a load on his head is doing work.
10. Work is
(a) Scalar quantity
(b) Vector quantity
(c) Neutral quantity
(d) F × S
11. A man pushes a wall but fails to displace it; he does
(a) negative work
(b) positive work
(c) maximum work
(d) No work at all
12. The work performed on an object does not depend on-
(a) applied Force
(b) displacement
(c) angle at which force is inclined to displacement
(d) initial velocity of the object.
13. A body carrying a box on his head is walking on level road from one place to another on a straight road is doing no work. This statement is -
(a) correct
(b) incorrect
(c) partly correct
(d) insufficient data
Answers:
1. b
2. d
3. d
4. b
5. a
6. a
7. d
8. d
9. d
10. a
11. d
12. d
13. a
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