A block whose mass is 1 kg is fastened

Asked by admin @ 17/08/2021 in Physics viewed by 294 People

A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N/m. 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 kinetic, potential and total energies of the block when it is 5 cm away from mean position

Answered by admin @ 17/08/2021

Answer:

K.E. = 0.19 J

P.E. = 0.0625 J

Total Energy = 0.25 J

Step by step explanation:

Angular frequency:

$\omega=\sqrt{k/m}\\ \omega=\sqrt{50/1}=7.07\:rad\:s^{-1}$

Its displacement at any time t is then given by:

$x(t)=0.1cos(7.07t)$

Therefore, when the particle is 5 cm away from the mean position (x=0.05m):

$0.05=0.1cos(7.07t)\\cos (7.07t) = 0.5\\sin (7.07t) = \sqrt{3/2}= 0.866$ .

Therefore,

$v(t)=\frac{dx(t)}{dt}\\v(t)=0.1*\omega sin(\omega t)\\v(0.05)=0.1*7.07*0.866=0.61\:m/s$

Hence the K.E. of the block= $\frac{1}{2}mv^2=\frac{1}{2}*1*(0.61)^2=0.19\:J$

The P.E. of the block = $\frac{1}{2}kx^2=\frac{1}{2}*50(0.05)^2=0.0625\:J\\$

The total energy of the block at x = 5 cm = K.E + P.E:

= 0.19 + 0.0625 = 0.25 J

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