Question

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

Answer 1

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

Answer 2

Answer:

Kinetic energy = 0.1875 J

Potential energy = 0.0625 J

Total energy = 0.25 J

Explanation:

K = 50 N/m      mass = 1 kg     Amplitude = 10 cm = 0.1 m

x = 5cm = 0.05 m

(i) Kinetic energy = 1/2 K ($A^{2}$ - $x^{2}$)

                             = 1/2 × 50 (0.1² - 0.05²)

                             = 0.1875 J

(ii) Potential energy = 1/2 K x²

                                 = 1/2 × 50 × 0.05²

                                 =0.0625 J

(iii) Total energy = Kinetic energy + Potential energy

                            = 0.1875 + 0.0625

                            =0.25 J