### Work

1. The meaning of work in our daily life is quite different than that in physics e.g., we may exert a large force on the wall, but if the wall remains intact in its position , then we have not done any work in the language of physics. Similarly, a get keeper performing his duty at the gate by standing there or a coolie taking a load on his head from one end to platform to the other are said to be doing work in the common language but in the language of physics either of them has not done any work.

2. In the language of physics, whenever force acting on a body is actually able to move it through some distance in the direction of force, then work is said to be done by the force. The force may act in any direction except perpendicular in the direction of displacement. Hence for work to be done, Following two conditions must be fulfilled:

a) A force must be applied

b) The applied force must produce a displacement in any direction except perpendicular to the direction of the force.

3. As the gate keeper standing at the gate is not involved in moving any load, hence both the force and displacement are zero here and so no work is done by him. Again, when a coolie moves on platform with load on his dead, he exerts force along the vertical direction. But because no distance is covered along the vertical direction, hence work done by the coolie is also zero.

### Work done by a constant force

1. The work done by a force is measured by the product of the applied force and the displacement of the body in the direction of the applied force.

If a force F acting on a body produces a displacement s in the body in the direction of the force , then the work done by the force is given by:

W = Fs …………(1)

Where F and s are the magnitude of the force vector and the displacement vector in the direction of force respectively.

2. If F does not act along the direction of displacement s (fig. 2), then work is calculated by the resolving the force F into two mutually perpendicular components:

i) F_{x} along the direction of displacement s(having magnitude F cosθ).

ii) F_{y} along the perpendicular to the displacement s (heaving magnitude F sinθ).

Where θ is the angle between the direction of force vector F and the displacement vector s.

Because the body does not move in the direction of F_{y}, hence work done in case of Component of F_{y} is zero. Hence, work is done only by component F_{x}, given by:

W = |F_{x}||s| = (F cos θ)s = F.s cosθ ………….(2)

Because F cos θ represents the components of applied force in the direction of displacement of the body, hence in general we can write:

Work = component of force in the direction of displacement × displacement

If we put θ = 0 in eq. (2), we have

W =F s cos 0° = Fs

Which is same as given by eq. (1), valid in case when F and s were in same direction. The eq.(2) is valid for all situations. Because F and s are vectors, hence eq.(2) can also be written as

W= F.s …………(3)

Hence, work done by a force represented as a dot product of the force applied and the displacement caused by the force. Because work done is the dot product of the vectors, it is a scalar quantity.

3. Also, if F and s are perpendicular to each other, then

W = F.s = F s cos 90° = 0

This means that if the displacement of the body is perpendicular to the force, no work is done, e.g. when a satellite revolves around the earth, the direction of force applied by the earth is always perpendicular to the direction of motion of the satellite. Hence work done on the satellite by the centripetal force is zero.

4. It should be noted here that in the formula for work done (W = F.s), force F remains constant both in magnitude and direction over the complete displacement s.

5. **Dimensions of work **: [W]=[F][s]=[MLT^{-2}][L]=[ML^{2}T^{-2}]