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Physics

Nine Standard >> Work

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Work and Gravitational Work

 

To perform a work some force should be applied. Without the application of force, no work will be done. The first condition of work done is to apply force. On the application of force if some displacement of the point of application of force takes place towards the direction of force then some work is done by the force. And work is measured by the product of force and displacement.

Work(W)=Force(F)  \(\times\) displacement(d)

In spite of the application of force, no work can be done if no displacement of the point of application of force. If force is applied in a heavy body but no displacement takes place in spite of applying a large quantity of force. Then word will be F \(\times\) 0= 0

W=F \(\times\) d \(\times\) \(\cos \theta\)
    [\(\theta\)=Angle between force and displacement]

Displacement should be towards the direction of the force. When displacement is done towards the direction of the force  

W= F \(\times\) d \(\times\) \(\cos 0^{o}\)
   = F \(\times\) d \(\times\) 1
   =F \(\times\) d

If you are walking with a suitcase in your hand on a horizontal path. The angle between force and displacement is \(90^{o}\). In this case, the work done will be

      W= F \(\times\) S \(\times\) \(\cos 90^{o}\)
          = F \(\times\) S \(\times\) 0
          =0
   
        If the force acts perpendicularly with the direction of displacement then no work is done despite occurring some displacement. This force is called no workforce.

Work Done in Favor of Force: Work done in favor of force refers to the work done when the force applied on an object and the displacement of the object occur in the same direction. This means that the force and displacement vectors are aligned, resulting in positive work being done.

Examples of Work Done in Favor of Force:

a) Pushing a box along a flat surface:

When you push a box horizontally, the force you apply and the displacement of the box occur in the same direction. Thus, the work done is positive.

b) Gravity pulling an object downward: When an object falls freely under the influence of gravity, the force of gravity and the displacement of the object are both downward.

Significance of Work Done in Favor of Force: Work done in favor of force represents the transfer of energy to an object. Positive work done implies that energy is being transferred to the object, resulting in an increase in its kinetic energy. This concept is fundamental in understanding the principles of energy conservation and the relationship between force and motion.

Work Done Against Force:

Work done against force refers to the work done when the force applied on an object and the displacement of the object occur in opposite directions. This means that the force and displacement vectors are anti-parallel, resulting in negative work being done.

  1. Examples of Work Done Against Force:

  2. a) Slowing down a moving object:

  3. When a force is applied in the opposite direction of an object's motion, the force and displacement vectors are anti-parallel. For instance, when applying brakes to a moving car, the force exerted by the brakes opposes the car's motion, resulting in negative work done against the force of motion.

  4. b) Lifting an object against gravity:

  5. When lifting an object against the force of gravity, the force applied and the displacement of the object are in opposite directions. In this case, work is done against the force of gravity, resulting in negative work.

  6. Significance of Work Done Against Force:

  7. Work done against force represents the transfer of energy from an object. Negative work done implies that energy is being removed or transferred away from the object, resulting in a decrease in its kinetic energy. Understanding this concept is crucial for analyzing systems where work is done against forces, such as braking mechanisms, resistance forces, or opposing gravitational forces

 
Unit of work:

1. CGS unit is dyne  \(\times\) cm= erg
2. SI unit is newton \(\times\) m=N.m= joule(J)
    1 J=1N \(\times\) m=\(10^{5}\) dyne \(\times\) \(10^{2}\) cm= \(10^{7}\) erg

Gravitational Unit:

1. 1 gram cm=980.6 erg
2. 1 kg m=9.8 J

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