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A 15.0-kg stone slides down a snow-covered hill (Fig.

ISBN: 9780321675460 31

Solution for problem 49P Chapter 7

University Physics | 13th Edition

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Problem 49P

A 15.0-kg stone slides down a snow-covered hill (Fig. P7.45), leaving point A at a speed of 10.0 m/s. There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.00 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively. (a) What is the speed of the stone when it reaches point B? (b) How far will the stone compress the spring? (c) Will the stone move again after it has been stopped by the spring?

Step-by-Step Solution:

Solution 49P Step 1: Data given Mass of stone m = 15.0 kg Velocity v = 10.0 m/s 1 Distance s = 100 m k = 2.00 N/m Height h = 20.0 m Kinetic coefficient C K 0.20 Static coefficient C = 0.80 S Step 2: In this question, we need to find velocity at point B This can be calculated using 2 2 v2 = v1 + 2ah Substituting values we get v 2 = v 2+ 2ad 2 1 v 2 = (10.0 m/s) + 2 × 9.8 m/s × 20.0 m 2 2 v2 = 492 v2= 22.18 m/s This can be approximated to 22.2 m/s Thus we have velocity at B as 22.2 m/s Step 3: We need to find the distance to which stone can compress the spring We shall find the kinetic energy of the stone It is obtained using 2 KE = 1/2 mv Substitute 2 2 m = 15.0 kg and v = v2 12+ v 22 = (10.0 m/s) + (22.2 m/s) Thus we get v = 493 m /s 2 2 KE = 1/2 × (15.0 kg × 493 m /s )2 2 KE = 4380 J Hence the kinetic energy of the rock when it hits the spring is 4380 J

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