11. Work and Energy-Magnitude of Work Science Class 9 In English-CBSE Notes
Last Updated : 22 March 2026
Explore science cbse notes for Class 9 Chapter 11. Work and Energy Topic Magnitude of Work in English medium for exams preparation.
11. Work and Energy-Magnitude of Work Science Class 9 In English-CBSE Notes
Last Updated : 22 March 2026
11. Work and Energy-Magnitude of Work Science Class 9 In English-CBSE Notes
11. Work and Energy
Magnitude of Work
Positive force and Negative force:
When we lift an object, the force we exert on the object will be considered positive. At the same time, there is another force working there which is called the force of gravity. The force of gravity works opposite to the force applied by us, so this force will be considered negative.
Since when we apply force on an object, we have to exert more force than the result of the force of gravity to displace the object, hence the resultant force becomes positive. For example, suppose we applied 20 N force to lift an object while the measure of gravity is 10 N there.
Resultant force = 20 - 10 = 10 N
In this situation, we applied a total of 10 N force to displace the object.
Example1: A person displaces a stone by 3 meters by applying 100 Newton force. So find out the work done by him.
solution:
Force applied here (F) = 100 N
Displacement (s) = 3 m
Work done (W) = F × s
= 100 × 3 = 300 joules
Example 2: A boy is unable to move a table by applying 20 N force and gets tired. So calculate the work done by him.
solution:
Force applied here (F) = 20 N
Displacement (s) = 0 m
Work done (W) = F × s
= 20 × 0 = 0 joules
The work done here is zero. Therefore, this work will not be considered.
Example 3: Suppose you lifted a heavy burden with force and placed it on your head. Displacement occurred in the load. Till this work was done, but if you keep this burden on your head and stand for a long time. So force is being felt by you, gravity is also working opposite to it, but there is no displacement in the object. Therefore, in this situation no work will be considered here.
Example 4: A porter carries a 25 kg load on his head 2 meters above. So calculate the work done by him on that burden.
solution :
Mass of burden m = 25 kg
Displacement = 2 m and
Force on object F = mg = 25 kg × 10 m s-2
= 250 kg / m s-2 or 250 N
Work on burden (W) = F × s
= 250 × 2 N m
= 500 N m = 500 J
1 joule work: When an object is displaced 1 meter in the direction of force by applying 1 N force, it will be said that 1 joule work has been done.
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NCERT Solutions for Class 9 science Chapter 11. Work and Energy Topic Magnitude of Work
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11. Work and Energy-Magnitude of Work Science Class 9 In English-CBSE Notes
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