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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 of `50 N m^(-1)`. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

Text Solution

Verified by Experts

Here `m=1 kg, k=50Nm^(-1), A=10cm=0.10m, y=5cm=0.05m`
Kinetic energy: `E_(K)=(1)/(2)k(A^(2)-y^(2))=(1)/(2)xx50[(0.10)^(2)-(0.05)^(2)]=0.1875J`
Potential energy, `E_(P)=(1)/(2)ky^(2)=(1)/(2)xx50xx(0.05)^(2)=0.0625J`
Total energy, `E=E_(K)+E_(P)=0.1875+0.0625=0.25J`
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