RCW7 solutions

# RCW7 solutions - Recitation ClassWork 7 solutions 1....

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Recitation ClassWork 7 solutions Physics 203 1. Chapter 10 Problem 29. REASONING AND SOLUTION If we neglect air resistance, only the conservative forces of the spring and gravity act on the object. Therefore, the principle of conservation of mechanical energy applies. When the 2.00 kg object is hung on the end of the vertical spring, it stretches the spring by an amount y, where 2 (2.00 kg)(9.80 m/s ) 0.392 m 50.0 N/m F mg y kk = = = = (10.1) This position represents the equilibrium position of the system with the 2.00-kg object suspended from the spring. The object is then pulled down another 0.200 m and released from rest ( 0 0 v = m/s). At this point the spring is stretched by an amount of 0.392 m + 0.200m = 0.592 m . This point represents the zero reference level ( 0 h = m) for the gravitational potential energy. h = 0 m: The kinetic energy, the gravitational potential energy, and the elastic potential energy at the point of release are: 22 11 0 KE (0m/s) 0J mv m = = = gravity PE (0m) mgh mg = = elastic 0 PE (50.0 N/m)(0.592 m) 8.76 J ky = = = The total mechanical energy 0 E at the point of release is the sum of the three energies above: 0 8.76 J E = . h = 0.200 m: When the object has risen a distance of 0.200 m h = above the release point, the spring is stretched by an amount of 0.592 m – 0.200 m 0.392 m = . Since the total mechanical energy is conserved, its value at this point is still 8.76 J E = . The gravitational and elastic potential energies are: 2 gravity PE (2.00 kg)(9.80 m/s )(0.200 m) 3.92 J mgh = = = elastic PE (50.0 N/m)(0.392 m) 3.84 J ky = = = Since gravity elastic KE PE PE E + += , gravity elastic KE – PE – PE 8.76 J – 3.92 J – 3.84 J 1.00 J E = = = h = 0.400 m: When the object has risen a distance of 0.400 m h = above the release point, the spring is stretched by an amount of 0.592 m – 0.400 m 0.192 m = . At this point, the total mechanical energy is still 8.76 J E = . The gravitational and elastic potential energies are:

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2 gravity PE (2.00 kg)(9.80 m/s )(0.400 m) 7.84 J mgh = = = 22 11 elastic PE (50.0 N/m)(0.192 m) 0.92 J ky = = = The kinetic energy is gravity elastic KE – PE – PE 8.76 J – 7.84 J – 0.92 J 0 J E = = = The results are summarized in the table below: h KE grav PE elastic PE E 0 m 0 J 0 J 8.76 J 8.76 J 0.200 m 1.00 J 3.92 J 3.84 J 8.76 J 0.400 m 0.00 J 7.84 J 0.92 J 8.76 J 2. Problem 82. Hint: Be really careful with part (a). What has to be true about the forces exerted on the box so that its acceleration is zero? REASONING a. The acceleration of the box is zero when the net force acting on it is zero, in accord with Newton’s second law of motion. The net force includes the box’s weight (directed downward) and the restoring force of the spring (directed upward). The condition that the net force is zero will allow us to determine the magnitude of the spring’s displacement.
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## This note was uploaded on 06/22/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).

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RCW7 solutions - Recitation ClassWork 7 solutions 1....

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