6. An ideal spring of un-stretched length 0.20 m is placed horizontally on a fri
ID: 2046298 • Letter: 6
Question
6. An ideal spring of un-stretched length 0.20 m is placed horizontally on a frictionless table. One end of the spring is fixed to a wall, and the other end is attached to a block of mass 8.0 kg. The 8.0 kg block is also attached to a string of negligible mass that passes over a frictionless pulley of negligible mass. A block of mass 4.0 kg hangs from the other end of the string.a. When the spring-and-blocks system is in equilibrium, the length of the spring is 0.25 m and the 4.0 kg block is 0.70 m above the floor.
i. Label the forces on each block.
ii. Calculate the tension in the string.
iii. Calculate the spring constant of the spring.
b. The string is now cut near the 8.0 kg block.
i. Calculate the time taken for the 4.0 kg block to hit the floor.
ii. Calculate the frequency of oscillation of the 8.0 kg block.
iii. Calculate the maximum speed attained by the 8.0 kg block.
Explanation / Answer
QUESTION 1: -------------------- For part (a), it's pretty easy. The large block has 4 forces acting on it: the force of the spring acting in the -x direction, the tension force in the string acting in the +x direction, the force of gravity in the -y direction, and the normal force acting in the +y direction. The small block has a tension force in the +y direction, and a gravitational force in the -y direction. (b) From the free body diagram for the small block, the tension in the string is found to be equal to the weight of the suspended block. Thus, T = m * g = 39.2 N (c) From the free body diagram of the large block, the force acting on the spring is equal to the tension in the string, which in (b) was found to be 39.2 N. Hooke's law can then be used to determine the force constant of the spring: F = k(x2-x1) = T k = T / (x2-x1) = 39.2 N / (0.25 - 0.20 m) = 784 N/m (d) When the string is cut, the small block accelerates towards the floor at a rate of 9/8 m/s^2 (gravitational acceleration). The relationship between time, distance and acceleration is: s = 1/2 * a * t^2 Solving for t: t = (2 * s / a)^(1/2) = (2 * 0.70 / 9.8)^(1/2) = 0.38 s (e) From the equations on page 3, you can calculate the period of the spring-mass system, then take the inverse to find its frequency: T = 2 * pi * sqrt(m / k) T = 0.635 s f = 1 / T = 1.576 Hz (f) Since the table is frictionless, the mass and spring will oscillate indefinitely, the energy of the system forever alternating between kinetic and potential (kinetic energy in the block, potential energy in the spring). The amplitude of the oscillations will be equal to the initial displacement of the block. Referring again to the equations provided, the potential energy stored in the spring at the instant the string is cut is: PE = 1/2 * k * x^2 = (1/2)(784)(.05)^2 = 0.98 J When all of this potential energy has been converted into kinetic energy, the velocity of the block will be at a maximum: KE = 1/2 * m * v^2 = 0.98 J Solving for v: v = sqrt(2(0.98 J) / (8kg)) = 0.495 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.