Work happens when a force moves an object over a distance. Imagine pushing a heavy box across the room – if you move it, you're doing work. But if you push against a wall and it doesn't move, no work is done. So, work is only done when there's both force and motion in the direction of that force.
We shall, in this document, understand how the scientific meaning of work is far different from the physiological pictures that come to our mind when we hear this word.
Consider the scenario of a person holding a weight at a height 'h' above the floor, as depicted in the figure.
In everyday language, we might say that the person is doing work, but according to our scientific definition, no work is performed by a force acting on an object at rest, possibly because the force applied by the person was not sufficient enough to produce any movement in the weight he was holding.
Say if a constant force F displaces a body through displacement s then the work done, W, is given by W =Fdcosθ =F.d, where d is the magnitude of displacement and θ is the angle between force and displacement. The SI unit of work is Joule or Newtonmetre.
In the S.I. system, the unit of work done is Joule. Another name for joule is newtonmeter.
In the 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 centimeter in its direction.
1 erg = 1 dyne × 1 cm = 1 g cm s^{2} × 1 cm = 1 g cm^{2} s^{2}
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
[Work] = [Force] [Distance] = [MLT^{2}] [L] = [ML^{2}T^{2}]
Looking at this equation W =Fdcosθ, we can understand three different scenarios for the work done:
Nature of Work
For a body freely falling under gravity work done by gravity is positive
An object is moved in a circular path by a string
Body lifted up
Special Cases
(i) If θ = 0°, W=FScos0° = FS
(ii) If θ =90°, W= FScos90° = 0
(iii) If θ =180°,W= FScos180° = −FS
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) Work done by gravitational force in the above case.
(c) Work done by friction on a body sliding down an inclined plane.
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Q1. A block of mass m = 2 kg is pulled by a force F = 40 N upwards through a height (h) = 2m. Find the work done on the block by the applied force F and its weight mg. (g = 10m/s^{2})
Solution:
Work done by the applied force W_{F}=Fh = cos0°
As the angle between force and displacement is 0°
W_{F}= (40)(2)(1)
= 80J
Similarly, work done by its weight
Q2. A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in fig. work done by normal reaction on the block in time t is:
(a)
(b)
(c) 0
(d)
Solution:
Here,
n = normal reaction
Now, work done by normal reaction 'N' on the block in time t,
or
Q3. A force is applied over a particle which displace it from its origin to the point The work done on the particle in joules is
(a) +10
(b) +7
(c) 7
(d) +13
Solution:
Given force
Q4. A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with horizontal. The coefficient of friction between the block and surface is μ. If the block travels with uniform velocity, find the work done by this applied force during a displacement d of the block.
Solution:
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
Q5. 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.
Solution:
(b) Work done by force,
= 10 + 0 + 0 + 6=16 N.m = 16 J
Q6. A block of mass m is placed on an inclined plane that is moving with constant velocity v in the horizontal direction as shown in the figure. Then find out the work done by the friction in time t if the block is at rest with respect to the incline plane.
Solution:
F.B.D of the block with respect to ground.
The block is at rest with respect to wedge
⇒ f = mg sinθ
In time t the displacement of the 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)
Up to this point, we have focused on the work performed by a constant force, maintaining both its magnitude and direction. Now, let's examine a force that consistently acts in a single direction but may vary in magnitude.
➢ When F as a function of x, y, z
Q7. A force F = (4.0 x + 3.0 y ) N acts on a particle that moves in the xdirection from the origin to x = 5.0 m. Find the work done on the object by the force.
Solution:
Here the work done is only due to the x component of force because displacement is along the xaxis.
Q8. 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.
Solution:
The 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)
When the force, denoted as F, is given as a function of time, denoted as t, it means that the magnitude of the force acting on an object can change over time. In this case, we are interested in finding the work done by this timedependent force in moving an object from one position to another.
Here is an example showing how to solve this:
Q9. 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, the block is at rest.
Solution:
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:
Work done by a force can be obtained from the area under the Fx graph. Unlike the integration
W = ∫Fdx= area under F x graph
Q10. Force acting on a particle varies with x as shown in the figure. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m.
Solution:
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.
Q11. A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m.
What is the work done by the person during the total movement of the box ?
(a) 3280 J
(b) 2780 J
(c) 5690 J
(d) 5250 J
Ans: (d)
Solution:
The given situation can be drawn graphically as
shown in figure.
Work done = Area under Fx graph
= Area of rectangle ABCD + Area of trapezium BCFE
Q12. Con sider a force The work done by this force in moving a particle from point A(1, 0) to B(0, 1) along the line segment is: (all quantities are in SI units)
(a) 2J
(b) 1/2J
(c) 1J
(d) 3/2J
Ans: (c)
Solution:
Q13. When a rubber band is stretched by a distance x, it exerts restoring force of magnitude F = ax + bx2 where a and b are constants. The work done in stretching the unstretched rubber band by L is:
(a) aL^{2} + bL^{3}
(b)
(c)
(d)
Ans: (c)
Solution:
Work done in stretching the rubber band by a distance dx is
dW = F dx = (ax + bx^{2})dx
Integrating both sides,
If the resultant or net force acting on a body is F_{net} then Newton's second law states that
F_{net} = ma ...(1)
If the resultant force varies with x, the acceleration and speed also depend on x.
Work done by net force F_{net} in displacing a particle equals to the change in kinetic energy of the particle.
i.e. we can write eq. (3) in following way
(W.D)_{C} + (W.D)_{N.C} + (W.D)_{ext.} + (W. D)_{pseudo}
= ΔK ...(4)
where (W.D)_{C} = work done by conservative force
(W. D)_{N.C} = work done by non conservative force.
(W.D)_{ext} = work done by external force
(W.D)_{pseudo} = work done by pseudo force, we know that
we know that
(W.D)_{c} =  ΔU
⇒  ΔU + (W.D)_{N.C} + (W.D)_{ext} (W.D)_{pseudo} = ΔK
= (W.D)_{N.C} + (W.D)_{ext} + (W.D)_{pseudo}  (k_{f} + u_{f})  (k_{i} + u_{i})
∴ k 4+ u = Mechanical energy.
=> work done by forces (except conservative forces) = change is mechanical energy.
If (W.D)_{N.C} = (W.D)_{ext} = (W.D)_{pseudo} = 0
= K_{f} + U_{f} = K_{i} + U_{i}
Initial mechanical energy = final mechanical energy
This is called mechanical energy conservation law.
Ex.14 The block shown in the figure is released from rest. Find out the speed of the block when the spring is compressed by 1 m.
Sol.
In the above problem only one conservative force (spring force) is working on the block so from mechanical energy conservation
at A block is at rest so k_{i} = 0
At position B if speed of the block is v then
Putting the above values in equation (i), we get
Ex.15 A block of mass m is dropped from height h above the ground. Find out the speed of the block when it reaches the ground.
Sol.
Figure shows the complete description of the problem only one conservative force is working on the block. So from mechanical energy conservation
k_{f} + u_{f} = k_{i} + u_{i}
(B) When two conservative force are acting in problem.
Ex.16 One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is attached to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the wall. Initially the spring makes an angle of 37º with the horizontal as shown in fig. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. [sin 37º = 3/5]
Sol. If l is the stretched length of the spring, then from figure
Now, taking point B as reference level and applying law of conservation of mechanical energy between A and B,
E_{A} = E_{B}
or
[as for, B, h = 0 and y = 0]
or
Ans.
Ex.17 The block shown in figure is released from rest and initially the spring is at its natural length. Write down the energy conservation equation. When the spring is compressed by l_{1 }?
Sol. Here two conservative forces are included in the problem.
(i) Gravitational force (ii) spring force
We assume zero gravitational potential energy at A as shown in figure.
from mechanical energy conservation
(C) When only one non conservative force is included in problem.
Ex.18 Find out the distance travelled by the block as shown in figure. If the initial speed of the block is v and m is the friction coefficient between the surface of block and ground.
Sol. Applying work energy theorem, we get
(D) When both conservative and nonconservative force in the problem
Ex.19 A particle slides along a track with elevated ends and a flat central part as shown in figure. The flat portion BC has a length l = 3.0 m. The curved portions of the track are frictionless. For the flat part the coefficient of kinetic friction is m_{k} = 0.20, the particle is released at point A which is at height h = 1.5 m above the flat part of the track. Where does the particle finally comes to rest?
Sol. As initial mechanical energy of the particle is mgh and final is zero, so loss in mechanical energy
= mgh. This mechanical energy is lost in doing work against friction in the flat part,
So, loss in mechanical energy = work done against friction
or mgh = mmgs i.e., s = = = 7.5 m
After starting from B the particle will reach C and then will rise up till the remaining KE at C is converted into potential energy. It will then again descend and at C will have the same value as it had when ascending, but now it will move from C to B. The same will be repeated and finally the particle will come to rest at E such that
BC + CB + BE = 7.5
or 3 + 3 + BE = 7.5
i.e., BE = 1.5
So, the particle comes to rest at the centre of the flat part.
Ex.20 A 0.5 kg block slides from the point A on a horizontal track with an initial speed 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/m. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. [g = 10 m/s^{2}]
Sol. As the track AB is frictionless, the block moves this distance without loss in its initial
KE = = × 0.5 × 3^{2} = 2.25 J. In the path BD as friction is present, so work done against friction
= m_{k} mgs = 0.2 × 0.5 × 10 × 2.14 = 2.14 J
So, at D the KE of the block is = 2.25  2.14 = 0.11 J.
Now, if the spring is compressed by x
0.11 = × k × x^{2}+ m_{k }mgx
i.e., 0.11 = × 2 × x^{2} 0.2 × 0.5 × 10x
or x^{2} + x  0.11 = 0
which on solving gives positive value of x = 0.1 m
After moving the distance x = 0.1 m the block comes to rest. Now the compressed spring exerts a force :
F = kx = 2 × 0.1 = 0.2 N
on the block while limiting frictional force between block and track is f_{L} = m_{s} mg = 0.22 × 0.5 × 10 = 1.1 N.
Since, F < f_{L}. The block will not move back. So, the total distance moved by block
= AB + BD + 0.1
= 2 + 2.14 + 0.1
= 4.24 m
(E) Important Examples :
Ex.21 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. Find the speed of the paritcle with respect to the sphere as a function of the angle q it slides.
Sol. We solve the above problem with respect to the sphere. So apply a pseudo force on the particle
Now from work energy theorem.
work done by ma = change in mechanical energy
⇒ ma R sinθ = (k_{f} + u_{f})  (k_{i} + u_{i})
maR sinθ = ⇒ = maR sinθ + mgR (1  cosθ)
⇒ v^{2} = 2R(a sinθ + g  g cosθ) ⇒ v = [2R (a sinθ + g  g cosθ)]^{1/2} m/sec
Ex.22 In the arrangement shown in figure m_{A} = 4.0 kg and m_{B} = 4.0 kg. The system is released from rest and block B is found to have a speed 0.3 m/s after it has descended through a distance of 1m. Find the coefficient of friction between the block and the table.
Neglect friction elsewhere. (Take g = 10 m/s^{2})
Sol. From constraint relations, we can see that
v_{A} = 2 v_{B}
Therefore, v_{A} = 2(0.3) = 0.6 m/s
as v_{B} = 0.3 m/s (given)
Applying W_{nc} = DU + DK
we get  m m_{A }g S_{A} =  m_{B} g S_{B} + m_{A}v_{A}^{2} + m_{B}v_{B}^{2}
Here, S_{A} = 2S_{B} = 2m as S_{B} = 1 m (given)
 m(4.0) (10) (2) =  (1) (10) (1) + (4) (0.6)^{2} + (1) (0.3)^{2}
or  80 m =  10 + 0.72 + 0.045 or 80m = 9.235 or m = 0.115 Ans.
The previous discussion holds good for this case, but instead of tension in the string we have the normal reaction of the surface. If N is the normal reaction at the lowest point, then the condition for the body to complete the circle holds for this case also. All other equations (can be) similarly obtained by replacing tension T by normal reaction N.
(B) When Body is attached to a rod of length R
In this case since the body is attached to a rigid rod. The body can not leave the circular path.
Therefore, if the speed of the body becomes zero before the highest point C. It's motion will be oscillatory about the centre of the rod.
Condition for completing the circle:
If the body just reaches the highest point then it will completes the vertical circle
Applying energy conservation between the lowest and highest point of circle, we get
So, If the velocity at point A is greater than equal to then body will compete the vertical circle.
(C) Vertical Motion in a Dual Ring
This system will behave as the preivious system. So u_{min} to complete vertical circle
Angle at which the normal reaction on the body will change its direction from inward to outward the ring is given by
(D) BODY MOVING ON A SPHERICAL SURFACE
The small body of mass m is placed on the top of a smooth sphere of radius R and the body slides down the surface.
At any instant, i.e., at point C the forces are the normal reaction N and the weight mg. The radial component of the weight is mgcos f acting towards the centre. The centripetal force is
The body flies off'the surface at the point where N becomes zero.
To find v, we use conservation of energy
From equation (ii) and (iii) we get
This gives the angle at which the body goes of the surface. The height from the ground of that point AD = R(1 + cos φ)
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1. What is work in physics? 
2. How is work done by a constant force calculated? 
3. What is the unit of work? 
4. How is work done by a variable force determined? 
5. What is the WorkEnergy Theorem in physics? 

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