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Question: Two springs of spring constants \(1000 \mathrm { Nm } ^ { - 1 }\)and \(2000 \mathrm { Nm } ^ { - 1 ...

Two springs of spring constants 1000Nm−11000 \mathrm { Nm } ^ { - 1 }and 2000Nm−12000 \mathrm { Nm } ^ { - 1 } are stretched with same force. They will have potential energy in the ratio of:

A

2 : 1

B

C

1 : 2

D

Answer

2 : 1

Explanation

Solution

Potential energy

U=12kx2=12k(Fk)2=F22k\mathrm { U } = \frac { 1 } { 2 } \mathrm { kx } ^ { 2 } = \frac { 1 } { 2 } \mathrm { k } \left( \frac { \mathrm { F } } { \mathrm { k } } \right) ^ { 2 } = \frac { \mathrm { F } ^ { 2 } } { 2 \mathrm { k } }

U1×k1=U2×k2\mathrm { U } _ { 1 } \times \mathrm { k } _ { 1 } = \mathrm { U } _ { 2 } \times \mathrm { k } _ { 2 } \quad or U1U2=k2k1=20001000=21\frac { \mathrm { U } _ { 1 } } { \mathrm { U } _ { 2 } } = \frac { \mathrm { k } _ { 2 } } { \mathrm { k } _ { 1 } } = \frac { 2000 } { 1000 } = \frac { 2 } { 1 }