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A block placed on a smooth horizontal floor is connected to a springs as shown. Initially an external force of 100N keeps it stretched by 1cm beyond natural lenjgth i equilibrium. This force is now removed and another force F is applied on this block, which slowly moves it from this position to a position where the spring is finally compressed by 3 cm. Find the work done by this force F (in joules).

Answer»


SOLUTION :Force CONSTANT `k=100/0.1 10^(4) N/m`
`W=1/2 k(x_("FINAL")^(2)-x_("initial")^(2))=4J`


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