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A simple harmonic oscillator consists of an 0.80 kg block attached to a spring (k = 200 N/m). The block slides on a horizontal frictionless surface about the equilibrium point x = 0 with a total mechanical energy of 4.0 J. (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete in 10 s? (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at x = 0.15 m?
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- A childs pogo stick (Fig. P7.69) stores energy in a spring with a force constant of 2.50 104 N/m. At position (x = 0.100 m), the spring compression is a maximum and the child is momentarily at rest. At position (x = 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 child and pogo stick is 25.0 kg. Although the boy must lean forward to remain balanced, the angle is small, so lets 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 childstickEarth system, taking both gravitational and elastic potential energies as zero for x = 0. (b) Determine x. (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 childs maximum upward speed. Figure P7.69arrow_forwardA lightweight spring with spring constant k = 225 N/m is attached to a block of mass m1 = 4.50 kg on a frictionless, horizontal table. The blockspring system is initially in the equilibrium configuration. A second block of mass m2 = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m2 loses contact with the springblock 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B). a. What is the subsequent speed of block 2? b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a). 77. (a) The energy of the system initially is entirely potential energy. E0=U0=12kymax2=12(225N/m)(0.150m)2=2.53J At the equilibrium position, the total energy is the total kinetic energy of both blocks: 12(m1+m2)v2=12(4.50kg+3.00kg)v2=(3.75kg)v2=2.53J Therefore, the speed of each block is v=2.53J3.75kg=0.822m/s (b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or v=0.822m/s. FIGURE P16.77arrow_forwardA horizontal spring attached to a wall has a force constant of k = 850 N/m. A block of mass m = 1.00 kg is attached to the spring and rests on a frictionless, horizontal surface as in Figure P7.55. (a) The block is pulled to a position xi = 6.00 cm from equilibrium and released. Find the elastic potential energy stored in the spring when the block is 6.00 cm from equilibrium and when the block passes through equilibrium. (b) Find the speed of the block as it passes through the equilibrium point. (c) What is the speed of the block when it is at a position xi/2 = 3.00 cm? (d) Why isnt the answer to part (c) half the answer to part (b)? Figure P7.55arrow_forward
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