Question

A block of mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 Nm. 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​

Answer 1

Explanation:

Answer:

K.E. = 0.19 J

P.E. = 0.0625 J

Total Energy = 0.25 J

Step by step explanation:

Angular frequency:

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

ω=

k/m

ω=

50/1

=7.07rads

−1

Its displacement at any time t is then given by:

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

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

\begin{lgathered}0.05=0.1cos(7.07t)\\cos (7.07t) = 0.5\\sin (7.07t) = \sqrt{3/2}= 0.866\end{lgathered}

0.05=0.1cos(7.07t)

cos(7.07t)=0.5

sin(7.07t)=

3/2

=0.866

.

Therefore,

\begin{lgathered}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\end{lgathered}

v(t)=

dt

dx(t)

v(t)=0.1∗ωsin(ωt)

v(0.05)=0.1∗7.07∗0.866=0.61m/s

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

2

1

mv

2

=

2

1

∗1∗(0.61)

2

=0.19J

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

2

1

kx

2

=

2

1

∗50(0.05)

2

=0.0625J

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

= 0.19 + 0.0625 = 0.25 J