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.

Explanation

The block executes SHM, its angular frequency, as given by Eq. (14.14b), is
`omega = sqrt((k)/(m))`
`= sqrt((50 "N m"^(-1))/("1kg"))`
`=7.07 "rad s"^(-1)`
Its displacement at any time t is then given by
`x (t)=0.1 cos (7.70 t)`
Therefore, when the particle is 5 cm away from the mean position , we have
`0.05 = 0.1 cos (7.07 t)`
Or `cos (7.07t)=0.5` and hence
`sin (7.07 t) = (sqrt(3))/(2)=0.866`
The velocity of the block at `x=5` cm is
`0.1 xx 7.07 xx 0.866 ms^(-1)`
`=0.61 ms^(-1)`
Hence the K.E of the block,
`= (1)/(2)mv^(2)`
`= 1//2 [1kg xx(0.6123 ms^(-1))^(2)]`
`=0.19 J`
The P.E. of the block
`= (1)/(2)kx^(2)`
`= 1//2 [50 "N m"^(-1) xx 0.05 m xx 0.05 m)`
`=0.0625 J`
The total energy of the block at `x = 5 cm`
`= K.E. + P.E`
`= 0.25 J`
We also know that at maximum displacement, K.E is zero and hence the total energy of the system is equal to the P.E Therefore, the total energy of the system,
`= 1//2[50 "N m"^(-1) xx 0.1 "m" xx 0.1 "m")`
`0.25 J`
which is same as the sum of the two energies at
a displacement of 5 cm. This is in conformity
with the principle of conservation of energy.