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A block whose mass is 1 kg is fastened t...

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 block when it is 5cm away from the mean position.

Text Solution

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Given , m = 1 kg, k = 50 N/m
A = 10 cm = 0.10 m
y = 5 cm = 0.05 m
Kinetic energy at a displacement y from the mean positions is
`E_(k) = (1)/(2)k(A^(2) - y^(2)) = (1)/(2) xx 50 xx [0.10)^(2) - (0.05)^(2)]`
= 0.1875 J
Potential energy is
`E_(p) = (1)/(2) ky^(2) = (1)/(2) xx 50 xx (0.05)^(2) = 0.0625` J
Total energy, `E = E_(k) + E_(p)`
0.25 J
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