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posted by  A-Hill on 11/1/2009 9:47:48 PM  |  status: Closed  |  Earned Karma: 32

Final Velocity with friction

Course Textbook Chapter Problem Needs by
Calculus Based Physics Physics for Scientists and Engineers 7th by Serway, Jewett 8 4P 11/1/2009 at 10:00:00 AM
Question Details:
A 10 kg object is moving at 6.36 m/s. The object passes  3m over rough surface with a frictional force of 20 N. The object then travels down a track and strikes a spring with a constant force of 2500N/m. How far does the object compress the spring. 
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posted by jerry2009 on 11/1/2009 9:55:23 PM  |  status: Live
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Response Details:


    Mass, M = 10 kg

    Initial velocity, U = 6.36 m/s

    Distance, S = 3 m

    Frictional force, f = 20 N

    Acceleration, a = - f / M   ( '-' sign represents retardation )

                           = - 20 / 10 = - 2 m/s^2
    Spring constant, K = 2500 N/m

    V2 - U2 = 2 a S

    V2 - 6.36 2 = 2 * - 2 * 3

    V = 5.33 m/s

    Kinetic energy of the object = Potential energy of the spring

    ( 1/2 ) M V 2 = ( 1/2 ) K X 2



             M V 2 =  K X 2

         10 * 5.33 2 = 2500 * X2

    Compression of the spring, X = 0.3371 m
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posted by Altair on 11/1/2009 10:01:46 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:
let the acceleration be a, we have:
F = -20N = ma
=>a = -20N/m = -20N/10kg = -2.0m/s2
vi = 6.36m/s,  distance s = 3m,  we need to find the vf, the speed after the 3m, we have:
vf2 - vi2 = 2as
=>vf = √[2as+vi2] = √[2*(-2.0m/s2)*3m + (6.36m/s)2 ] = 5.33 m/s
the kinetic energy after the rough surface is:
KE = (1/2)*10kg*(5.33m/s)2 = 142.04 J
all these energy will be converted to the potential energy of the spring:
PE = (1/2)k (Δx)2 = KE = 142.04J
=>Δx = √[2*142.04J/2500N/m] = 0.34 m

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