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A spring of spring constant 5 × 103 N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm isa) 12.50 N-mb) 18.75 N-mc) 25.00 N-md) 6.25 N-m
Here, W=1/2k(x22-x12) = 1/2×5×103((10/100)2-(5/100)2) =1/2×5×103/104(100-25) =18.75JHence the work required to stretch the spring further by another 5cm is 18.75 N-mThanks and regards
Work done by spring=1/2k(x22-x12)K=5×103N/mX2=10×10-2mX1=5+10-2mW=1/2×5×103(100-25)10-4W=18.75N-m
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