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Get Full Access to University Physics - 13 Edition - Chapter 6 - Problem 68p
Get Full Access to University Physics - 13 Edition - Chapter 6 - Problem 68p

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# A 5.00-kg package slides 1.50 m down a long amp that is ISBN: 9780321675460 31

## Solution for problem 68P Chapter 6

University Physics | 13th Edition

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Problem 68P

A 5.00-kg package slides 1.50 m down a long amp that is inclined at 24.0° below the horizontal. The coefficient of kinetic friction between the package and the ramp is ???k = 0.310 Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 m/s at the top of the ramp, what is its speed after sliding 1.50 m down the ramp?

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Solution 68P Mass of the package is 5 kg. The length of the inclined plane is 1.5 meters. 0 The angle of inclination is 24 . Coefficient of kinetic friction is, k = 0.310. a) The frictional force is, F fric= kN = 0.310 × mg = 0.310 × 5 × 9.8 = 15.19 Newton The direction of frictional force is antiparallel to the direction of displacement. So, the work done by the frictional force is, 0 W fric F fricdS = F fric× S × cos 180 = 15.19 × 1.5 × ( 1) = 22.785 jules. The minus sign indicates the work done is negative. b) The gravitational force is acting in the downward direction. F = mg = 5 × 9.8 = 49 N . gravity The displacement is 1.5 meters. The angle between the force and displacement vector is, = 180 (90 + 24) = 180 114 = 66 . The work done by the gravitational force is, W = F. dS = | || |s gravity 0 = 49 × 1.5 × cos 66 = 30 jules. c) The normal force is, N = mg cos 24 = 5 × 9.8 × cos 24 = 44.76 N . The normal force and displacement are perpendicular to each other. So, the work done by the normal force would be, 0 W normal= F normal. dS = 44.76 × 1.5 × cos 90 = 0jules. There is no work done by the normal force. d) The total work done is, W total 22.785 + 30 + 0 = 9.215 jules. e) If the package has a speed of 2.2 m/s at the top of the ramp, the kinetic energy would be, K.E = 1/2 × m × v = 1/2 × 5 × 2.2 = 12.1 jules. initial The net work done we found as, W = 9.215 jules. From work-energy theorem we know that, w = K.E final K.E initial 9.215 jules = 1/2 × m × V 2 12.1 f 1/2 × m × V 2 = 9.215 + 12.1 = 21.315 f 2 v f = (21.315 × 2) / 5 = 8.526 v = f8.526 = 3 m/s The final velocity would be 3 m/s.

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##### ISBN: 9780321675460

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