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Question

A spring of spring constant 5 x 103 N/m is etched initially by 5 cm from the unstretched position the work required to stretch it further by another 5 cm is
  1. 18.75Nm
  2. 25.00Nm
  3. 6.25Nm
  4. 12.50Nm

A
25.00Nm
B
12.50Nm
C
18.75Nm
D
6.25Nm
Solution
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Work done is change in potential energy
The spring is initially stretch to 5 cm and again further stretched by 5 cm and
further stretched by 5 cm
w=p10p5=12×5×103[I(10100)2(5100)2]=12×5×103(10100+5100)(101005100)Nm=12×5×103×15100×5100=2.5×15×5×103×104=75×2.5×101=7.5×2.5Nm=18.75Nm
Option A.

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