➢ Work done by a force is zero if displacement is perpendicular to the force (q = 90°)
Example.
The tension in the string of a simple pendulum is always perpendicular to displacement. (Figure)
So, work done by the tension is zero.
➢ If the angle between force and displacement is acute (q < 90°), we say that the work done by the force is positive.
Example:
When a load is lifted, the lifting force and the displacement act in the same direction. So, work done by the lifting force is positive.
Example:
When a spring is stretched, both the stretching force and the displacement act in the same direction. So work done by the stretching force is positive.
➢ If the angle between force and displacement is obtuse (q > 90°), we say that the work done by the force is negative.
Example:
When a body is lifted, the work done by the gravitational force is negative. This is because the gravitational force acts vertically downwards while the displacement is in the vertically upward direction.
Fig: Forces acting on a box.
In cgs system, the unit of work is erg.
One erg of work is said to be done when a force of one dyne displaces a body through one centimetre in its own direction.
1 erg = 1 dyne × 1 cm = 1 g cm s^{2} × 1 cm = 1 g cm^{2} s^{2}
Note : Another name for joule is newton metre.
Relation between joule and erg:
1 joule = 1 newton × 1 metre
1 joule = 10^{5} dyne × 10^{2} cm = 10^{7} dyne cm
1 joule = 10^{7} erg
1 erg = 10^{7} joule
➢ Dimensions of Work
[Work] = [Force] [Distance] = [MLT^{2}] [L] = [ML^{2}T^{2}]
Work has one dimension in mass, two dimensions in length and `2' dimensions in time, On the basis of dimensional formula, the unit of work is kg m^{2} s^{2}.
Note that 1 kg m^{2}_{ }s^{2} = (1 kg m s^{2}) m = 1 N m = 1 J.
If several forces act on a particle, then we can replace in equation by the net force where
This gives the work done by the net force during a displacement of the particle.
We can rewrite equation (i) as :
So, the work done on the particle is the sum of the individual work done by all the forces acting on the particle.
Q.1. A block of mass M is pulled along a horizontal surface by applying a force at an angle q with horizontal. Coefficient of friction between block and surface is m. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block.
Ans: The forces acting on the block are shown in Figure. As the block moves with uniform velocity the resultant force on it is zero.
F cos θ = μN ...(i)
F sin θ + N = Mg ...(ii)
Eliminating N from equations (i) and (ii),
F cos θ = n(Mg  Fsin θ)
Work done by this force during a displacement d
Q.2. A particle moving in the xy plane undergoes a displacement while a constant force acts on the particle.
(a) Calculate the magnitude of the displacement and that of the force.
(b) Calculate the work done by the force.
Ans:
(b) Work done by force,
= 10 + 0 + 0 + 6=16 N.m = 16 J
Q.3. A block of mass m is placed on an inclined plane which is moving with constant velocity v in horizontal direction as shown in figure. Then find out work done by the friction in time t if the block is at rest with respect to the incline plane.
Ans: F.B.D of block with respect to ground.
Block is at rest with respect to wedge
⇒ f = mg sinθ
In time t the displacement of block with respect to ground d = vt
Work done by friction for man A
W_{f} = (component of friction force along displacement) × displacement
W_{f} = mgsinθ.vt cos(180° θ)
W_{f} =  mg vt cosθ sinθ
W_{f} for man B = 0 (displacement is zero with respect to man B)
➢ When F as a function of x, y, z
Q.4. A force F = (4.0 x + 3.0 y ) N acts on a particle which moves in the xdirection from the origin to x = 5.0 m. Find the work done on the object by the force.
Ans. Here the work done is only due to x component of force because displacement is along xaxis.
Q.5. A force F = 0.5x + 10 acts on a particle. Here F is in newton and x is in metre. Calculate the work done by the force during the displacement of the particle from x = 0 to x = 2 metre.
Ans. Small amount of work done dW in giving a small displacement is given by
Total work done, W
➢ When F is given as a function of Time(t)
Q.6. The force F = 2t^{2} is applied on the 2 kg block. Then find out the work done by this force in 2 sec. Initially at time t = 0, block is at rest.
Ans.
Let the displacement of the block be dx from t = t to t = t +dt then, work done by the force F in this time interval dt is.
➢ Area Under Force Displacement Curve
Q.7. Force acting on a particle varies with x as shown in figure. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m.
Ans: The work done by the force is equal to the area under the curve from x = 0 to x = 6.0 m.
This area is equal to the area of the rectangular section from x = 0 to x = 4.0 m plus the area of the triangular section from x = 4.0 m tox = 6.0 m. The area of the rectangle is (4.0) (5.0) N.m = 20 J, and the area of the triangle is (2.0), (5.0) N.m = 5.0 J. Therefore, the total work done is 25 J.
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