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# A small glider is placed against a compressed spring at

ISBN: 9780321675460 31

## Solution for problem 45E Chapter 6

University Physics | 13th Edition

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Problem 45E

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 40.0o above the horizontal. The glider has mass 0.0900 kg. The spring has k = 640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Step-by-Step Solution:

Solution 45E Step 1: Provided that, the glider and spring system makes an angle of 40° with the horizontal. The distance moved, S = 1.80 m According to the work-energy theorem, work done = Change in kinetic energy At maximum point, the KE will be completely PE. Step 2: According to work-energy theorem, kinetic energy+Potential energy = total energy ½ mv + ½ kx = Total energy 2 Therefore, at maximum point, ½ kx = total energy (v = 0 at maximum height) 2 ½ kx = mg (h sin40) Provided, k = 640 N/m ½ ×640 N/m × x = 0.0900 kg × 9.8 m/s × 1.8 m × 0.643 (where, sin 40 = 0.643) Therefore, we can write, 2 x = 1.02 J × 2 / k Therefore, x = 2.04 J / 640 N/m = 0.0032 m 2 Taking square root on both sides, x = 0.0564 m = 5.64 cm The spring was originally compressed to 5.64 cm. Step 4: b) At 0.8 m, the spring will not have any contact with the glider because, it will come back to its initial position at 5.64 cm. The potential energy at this point will be, PE = mg hsin 40 2 That is, PE = 0.0900 kg × 9.8 m/s × 0.8 m × 0.643 = 0.454 J Therefore, according to the work-energy principle, W = PE + KE Or, KE = W - PE W = mgh = 0.0900 kg × 9.8 m/s × 1.8 m × 0.643 = 1.02 J KE = 1.02 J - 0.454 J = 0.566 J At 0.8 m, the KE of the glider will be, 0.566 J

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##### ISBN: 9780321675460

The answer to “A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 40.0o above the horizontal. The glider has mass 0.0900 kg. The spring has k = 640 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?” is broken down into a number of easy to follow steps, and 124 words. This textbook survival guide was created for the textbook: University Physics, edition: 13. The full step-by-step solution to problem: 45E from chapter: 6 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 45E from 6 chapter was answered, more than 965 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: spring, glider, track, distance, air. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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