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A block whose mass is 1 kg is fastened to a spring.The spring has a spring constant `50Nm^(-1)`. The block is pulled to a distance `x=10cm` from its equilibrium position at `x=0` on a frictionless surface at `t=0`. Calculate the kinetic, potential and total energies of the blocak when it is 5cm away from the mean position. |
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Answer» Since, time is noted from the equilibrium position for SHM, hence displacement, x at time t is `x=x_(0)sinomegat,` where ` omega=sqrt((k)/(m))=sqrt((50)/(1))Nm^(-1)` Velocity, `V=(dx)/(dt)=x_(0)omega cosomegat` When `x=5cm=0.50m` and `x_(0)=10cm=0.10m` Then, `0.05=0.10sinomegat` or `sinomegat=(1)/(2)=sin((pi)/(6))` or `omega=(pi)/(6)` `:. cosomegat=cos((pi)/(6))=(sqrt(3))/(2)` Hence, `V=0.1XXsqrt(50)xxcos((pi)/(6))` `=0.1xxsqrt(50)xx(sqrt(3))/(2)=0.6123m//s` K.E. of block, `E_(K)=(1)/(2)mV^(2)` `=(91)/(2)xx1xx(0.6123)^(2)=0.1875J` P.E. of block , `E_(P)=(1)/(2)kx^(2)` `=(1)/(2)xx50xx(0.05)^(2)=0.0625J` Total energey `=E_(K)+E_(P)=0.875+0.0625` `=0.25J` |
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