An ideal spring of outstretched length 0.20 m is placed horizontally on a fricti

Question

An ideal spring of outstretched length 0.20 m is placed horizontally on a frictionless table. One end of the spring is fixed and the other end is attatched to a block of mass M= 8.0 kg. The 8.0 kg block is also attatched to a massless string that passes over a small frictionless pulley. A block of mass m= 4.0 kg hangs from the other end of the string. When this spring-and-box system is at equilibrium, the length of the string is 0.25 m and the 4.0 kg block is 0.70m above the floor. A) Calculate the tension in the string and the force constant of the spring. The string is now cut at point P. B) Calculate the time take by the 4.0 kg block to hit the floor. C) Calculate the frequency of oscillation of the 8 kg block D) Calculate the maximum speed attained by the 8 kg block

Explanation / Answer

The system is initially in equilibrium, so the forces on each mass will be also in equilibrium. Though YOU will have to draw the free body diagrams here are some tips: 8 kg block: Vertical forces are (a) its weight W1 = - m1 g going downward and (b) the table's reaction: R1 = -W1 = m1 g Horizontal forces are: a) the restitution force exerted by the spring (if we imagine x axis growing to the right and the fixed end of the spring at the left, we'll consider this force negative as opposed to possitive displacements)... F = - k x k = spring constant x = the displacement from the neutral length = L1-Lo (length when stretched minus original or neutral length) b) the force exerted by the string due to the weight of the 4 kg mass. The pulley only deviates the direction of this force and produces no changes in its module: W2 = m2 g = - F (though with our convention a downward force is negative, by changing direction and being referred to the horizontal axis it is positive in this position). 4 kg block We only have vertical forces: W2 = -m2 g weight, downward T2 = -W2 = m2 g string tension, upward (T2 = tension in the string) Calculation of the tension in the string: ------------- ------------ ---------------- ------------- T2 = m2 g = 4kg 9.8 m/s² = 39.2 N - - - - - - - - - - - - - - - - - - - - - - - - - - - - Calculation of the force constant of the spring: ------------ ---------------- --------------- -------------------- From the horizontal forces equilibrium for the 8kg block: k (0.25m - 0.20m) = m2 g = T2 = 39.2 N k = 39.2N / 0.05 m = 784 N/m - - - - - - - - - - - - - - - - - - - - - - - - Frequency of oscillation ---------------- ------------------ Once the string is cut between the 8kg block and the pulley, it oscillates with an angular frequency ?: ? = v(k / m1) = v( 784 kg m/s² / 8 kg) = 9.9 / s ? = 2 ¶ f => f = ? / 2¶ = 1.58 Hz - - - - - - - - - - - - - - - - - - - - - - - - - - where f is the requested frequency of oscillation. Maximum speed of m1 -------------- ----------------- The position of m2 will be described by: x = 0.05m cos w t being t=0 when the block is initially in equilibrium and starts the oscillation. => xmax = 0.05m cos 0 = 0.05m which we already knew. The speed is: v = dx/dt = - w xmax sin w t where the maximum speed is its module when sin wt=1: vmax = w xmax = 9.9/s * 0.05m = 0.495 m/s - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Time in which m2 to hits the floor: --------------------- ------------------------ h2 = 0.7m hf = 0m (h @ floor level or final h) v2o = 0 in equilibrium at 0.7m t = requested time hf = 0 = h2 + v2o t - 1/2 g t² h2 = 1/2 g t² finally: t = v(2 h2 / g) = v(1.4 m / 9.8 m/s²) = 0.38 s - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Summary of results =============== Tension = T2 = 39.2 N Constant = k = 784 N/m frequency = f = 1.58 Hz vmax = 0.495 m/s t = 0.38s Hope this helps.

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