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A particle move in a straight line with retardation proportional to its displacement its loss of kinectic energy for any displacement `x` is proportional to |
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Answer» Here, mass`=`m. Retardation `(-a)` of particle is proportional to displacement `(x)` `:. -apropx or a=-kx` ..(i) Now, `a=(dv)/(dt)=(dv)/(dx).(dx)/(dt)=(dv)/(dx)(v)` From (i), `v(dv)/(dx)=-kx` or `v dv=-kx dx ` …(ii) Suppose velocity of particle changes from u to v during its displacement x. `:.` Intefrating (ii), within proper limits, `int_(u)^(v)vdv=-kint_(0)^(x)xdx` `[(v^(2))/(2)]_(u)^(v)=-k[(x^(2))/(2)]_(0)^(x)` `(1)/(2)v^(2)-(1)/(2)u^(2)=-k(x^(2))/(2)` or `(1)/(2)u^(2)-(1)/(2)v^(2)=(1)/(2)kx^(2)` Loss in KE`=(1)/(2)m u^(2)-(1)/(2)mv^(2)=(1)/(2)mkx^(2)` |
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