Question

3. A 1.2 kg block sliding at 6.0 m/s on a frictionless surface runs into and sticks to a spring. The spring is compressed 0.10 m before stopping the block and starting its motion back in the opposite direction. a. Calculate the energy of vibration. b. Calculate the force constant of the spring. c. Calculate the period of vibration of the system. d. Write the equation of motion of the block when its oscillating.

Answer 1

(a) Vibration Energy = 1/2 mv^2 = 1/2 times 1.2 times 6^2 E = 0.6 times 36 E = 21.6 Joule (b) 1/2 kx^2 = 1/2 mv^2 spring potential Energy = kinetic Energy x = 0.1 m v = 6 m/s m = 1/2 kg k times 0.1^2 = times 1.2 times 6^2 k = 4320 N/m (c) Period of motion R = 2 phi squareroot m/k T = 2 phi squareroot 1.2/4320 T = 0.105 second (d) x = A sin (wt) w = squareroot k/m = minus 0.1 sin (squareroot k/m t + phi/2) x = minus 0.1 sin (60 t + phi/2) x = minus 0.1 cos (60 t)