The potential energy of a certain spring when stretched through a distance s is 10 j.The amount of work done is that must be done on this spring
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Given P.E to stretch throgh 'x' is
1212kx2=10Jkx2=10J
To stretch through additonal distance 'x'
1212k[(2s)2−s2]=3×12k[(2s)2−s2]=3×12×ks2×ks2
=3×10J=3×10J
=30J
1212kx2=10Jkx2=10J
To stretch through additonal distance 'x'
1212k[(2s)2−s2]=3×12k[(2s)2−s2]=3×12×ks2×ks2
=3×10J=3×10J
=30J
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