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CBSE Class 11-science Answered

Body A is moved to x = -5 cm and body B is moved to x = 15 cm. If force constant is 100 N/m, find the restoring force on both A(left body) and B(right body).
Asked by bjayanta | 14 Aug, 2017, 04:11: PM
answered-by-expert Expert Answer
- Restoring force for body A on left will be towards right and similarly towards the left for body B (which is on the right) .
- As restoring force is a property of spring, it will be same for both the bodies.
- Considering , the given diagram shows equilibrium position of the spring, the unstretched length of the spring is 5 cm.
- When A is moved 5 cm in the left direction, it is given that B moves 15 cm to the right.
- Therefore, the 'x' of a spring, that is the length by which the spring is stretched with respect to its equilibrium is 5 cm (on the left) plus 15 cm - 5 cm (towards the right) = 10 cm, that is k=10cm.
- The formula for Restoring force (F)=- kx
- Here, k is the spring constant and 'x' is the length of the spring to which it is stretched with respect to its equilibrium position. 
Substituting the given values in the formula,    
F = -(100 x10) = -1000 dyne/cm
The negative symbol indicates that the force is directed opposite to the dircetion of movement of the bodies attached to the spring. 
Answered by Abhijeet Mishra | 21 Aug, 2017, 04:07: PM

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