A 2 kg block is dropped from a height 0.4m on a spring of force constant k=1960N/m.The maximum compression of spring is
0.1m
0.2m
0.3m
0.4m
A
0.1m
B
0.3m
C
0.4m
D
0.2m
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Solution
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By law of conservation of energy , we have loss in gravitational potential energy=gain in elastic potential energy ⇒mg(h+x)=12kx2 2×10(0.4+x)=12×1960×x2 ⇒8+20x=980x2 ⇒980x2−20x−8=0 Solving ,we get x=0.1m
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