a stretching force of 1000N is applied at one end of a spring balance and an equal stretching force is applied at the other end at the same time.what will be the reading of the balance

Dear Student ,
Here in this case the force which is directly applied on the spring will produce stretching of the spring .
Therefore ,F=-kxx=-Fk=-1000kunithere k is the force constant of spring .
Regards

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