44.the spring constant of two spring are k and k' respectively Spring are stretch up to the limit when potential energy of both becomes equal the ratio of applied force( F1 and F2 )on them will be

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Q. The spring constant of two springs are

K_{1}

and

K_{2}

respectively springs are stretch up to that limit when potential energy both becomes equal. The ratio of applied force

(F_{1}

and

F_{2})

on them will be:

Two springs have their force constants $K_{1}$ and $K_{2}$ . Both are stretched till their elastic energies are equal. Then, ratio of stretching forces $F_{1} / F_{2}$ is equal to :

Q. Two springs have their force constants

K_{1}

and

K_{2}

. Both are stretched till their elastic energies are equal. If the stretching forces are

F_{1}

and

F_{2}

then

F_{1} : F_{2}

is equal to

Two springs of spring constants 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have potential energy in the ratio

Q. Two springs of spring constants

1500 N / m

and

3000 N / m

respectively are stretched with the same force. They will have the potential energies in the ratio:

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