A child\'s pogo stick as shown below stores energy in a spring with a force cons

Question

A child's pogo stick as shown below stores energy in a spring with a force constant of 2.55 104 N/m. At position (xA = -0.150 m), the spring compression is a maximum and the child is momentarily at rest. At position (xB = 0), the spring is relaxed and the child is moving upward. At position , the child is again momentarily at rest at the top of the jump.
The combined mass of the child and the pogo stick is 28.5 kg. Although the boy must lean forward to remain balanced, the angle is small, so let's assume the pogo stick is vertical. Also assume the boy does not bend his legs during the motion.
(a) Calculate the total energy of the child-stick-Earth system, taking both gravitational and elastic potential energies as zero for x = 0.
(b) Determine xC
(c) Calculate the speed of the child at x = 0.
(d) Determine the value of x for which the kinetic energy of the system is a maximum.
(e) Calculate the child's maximum upward speed.

Explanation / Answer

The spring constant, k = 2.55 * 10^4 N/m The mass ( pogo stick + child ) = 28.5 kg The compression in the spring, x = 0.15 m a) The potential energy, U = mgh = -28.5 * 9.8* 0.15 = -41.9 J     The spring energy, E = (1/2)kx^2 = 0.5* 2.55 * 10^4 * 0.15*0.15 = 286.9 J     So the total energy at x = 0 is = U + E = -41.9 + 286.9 = 245 J b) To solve for the value of xC we can equate the spring energy with the gravitational energy. mgxC = (1/2)kx^2                 xC =  (1/2)kx^2 / mg =( 286.9/28.5*9.8) = 1.03 m. c) According to law of conservation of energy the total energy at x = 0 is converted into kinetic energy when it is in motion. So, TE = KE TE = 0.5mv^2 From the above, v = sqrt(TE/0.5m)^2 = 4.15 m/s d) From the law of conservation law of enrgy 0.5kx^2 + mgx = 245 J By substituting and simplifying the above equation we get x = 0.13 So the maximum comprission in the spring = 0.13 m e) any where the net force is zero. So, 0.5mv^2 + mgx + 0.5kx^2 = 0 By differentiating and simplifying the above equation we get the value of maximum upward speed.

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