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NEET-XII-Physics

08: Work and Energy

with Solutions - page 2
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  • #2
    Two springs A and B(kA = 2kB) are stretched by applying forces of equal magnitudes a the four ends. If the energy stored in A is E, that in B isA
    (a) E/2
    (b) 2E
    (c) E
    (d) E/4
    digAnsr:   b
    Ans : (b) 2E
    Let xA and xB be the extensions produced in springs A and B, respectively.
    Restoring force on spring A, `` F={k}_{\,\mathrm{\,A\,}}{x}_{\,\mathrm{\,A\,}}`` ...(i)
    Restoring force on spring B, `` F={k}_{\,\mathrm{\,B\,}}{x}_{\,\mathrm{\,B\,}}`` ...(ii)
    From (i) and (ii), we get:
    `` {k}_{\,\mathrm{\,A\,}}{x}_{\,\mathrm{\,A\,}}={k}_{\,\mathrm{\,B\,}}{x}_{\,\mathrm{\,B\,}}``
    It is given that kA = 2kB
    `` \therefore {x}_{\,\mathrm{\,B\,}}=2{x}_{\,\mathrm{\,A\,}}``
    Energy stored in spring A:
    `` E=\frac{1}{2}{k}_{\,\mathrm{\,A\,}}{{x}_{\,\mathrm{\,A\,}}}^{2}`` ...(iii)
    Energy stored in spring B:
    `` E\text{'}=\frac{1}{2}{k}_{\,\mathrm{\,B\,}}{{x}_{\,\mathrm{\,B\,}}}^{2}=\frac{1}{2}\left(\frac{{k}_{\,\mathrm{\,A\,}}}{2}\right)(2{x}_{\,\mathrm{\,A\,}}{)}^{2}``
    `` \therefore E\text{'}=2\times \left(\frac{1}{2}{k}_{\,\mathrm{\,A\,}}{{x}_{\,\mathrm{\,A\,}}}^{2}\right)=2E[\,\mathrm{\,From\,}(\,\mathrm{\,iii\,}\left)\right]``
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