Question

Question 4 25 pts A box of mass of 3 Kg is attached to a spring with spring constant of 20 N/m. A force F= 5N is applied to pull the spring right. 1. Draw the diagram and label it correctly. 2. How much does the spring stretch? 3. What is the elastic potential energy stored in the spring after it is pulled out. 4. Assuming, surface is frictionless, What will be the speed with which box moves past its equilibrium position. 5. What will be the period and frequency of the Oscillation.

Answer 1

Ans:

Ans: Given m = 3kg k = 20 N/m F = 5N x is extension or Stretch in spring. Didgram x=0 at equilibrium. k=20 Nm prorogoro (when no force is applied ) A sro F =5N kx (when force F is applied) 2 Now law g acc. to hookes force applied on spring can be given by, F = -KX (-ve sign shows that it is in oppo.direction). 5 = j F = kx 5 - B 20 (x) = 0.25 4 The spoing will stretch upto 0,25 m 0.25m due to force F=5N 20 A*

9) The Elastic potential energy stored after it is pulled out is given as, Spring ofter P.E. = I K (x2 2 1 x 20 0.25,2 2 0.625 Nom or J *** PE= 0.625 I 4) 7. 지기 * * Totol Energy is given by E = Find (force x extension) which is addition of P. E. & K. E KRE= I mra & P.E. = 1 kx2 2 2 *** frictionless Assuming slf is E = P. E + K. E E PE P.Eode KE = Im V2 = 2 Ikx2 2 --- (5) 0.25) 1 (20)(0.25)2 1-25 0.625 Imrza 2 V2 - 2 x (0.625) 1025 3 3

~2 = 0.4166 v= 0.6455 mis .speed with which box moves past its eq v=0.6455 mis m is 5) → period of Oscillation is given by T = 21 к = 2T 3 T = 2.4334 sec cJ *** period of oscillation is 2.4334 sec. frequency of oscillation is g)ven by, f = 1 ) ㅜ 2.4334 If= 0.4 109 Hz or /sec. I frequency of Oscillation = 0.4109 Hz

The stretch in the spring = 0.25 m.

Elastic potential energy stored in the spring = 0.625 Joules.

Speed with which the box moves past the equilibrium position = 0.6455 m/s.

Period of oscillation = 2.4334 seconds .

Frequency of oscillation = 0.4109 Hz.