Question

Chapter 07, Problem 44 GO A 70.0-kg person, running horizontally with a velocity of +2.49 m/s, jumps onto a 19.3-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow? (a) Number (b) Number Units Units

Answer 1

given: mass of person m, = 70 kg Initial velocity of person vi = 2.49 m/s mass of seed M2 = 19.3 kg initial velocity of sled= o Cat rest) Let the final velocity of person with sled = V2 (a) From the conservation of momentum m. v, + m2 Xo = (m, + M2)x V2 70x 2.49 = (70 + 19.3) *V2 174 3 = 89.3 V2 V₂ = 1.95 mis the velocity of sled & person (6) For the mation of person with sled Initial velocity = 1.95 m/s distance travelled a=30m final velocity = 0 Energy lost in friction = kinetic energy of the person with sled Ef (m+m2) xV2 = 4(89.3) X(1.95)2 Ef = 169.78 T work done by frictional force = Energy lost in friction Wf = Ef Frictiona fora x distance moved a 169.78 - Fix d = 169.78 *** 169.78/30 - 5.66 N Frictional force Ff = 5.66 N.

onsider the EBD of person motion 4R with sled during the E Y Lux + m + M₂) g Since no mation in vertical direction { fy=0 R=(m + m₂) = (70+ 19.3).g R= 89.3x9.8 IR = 875.14 N. Frictional force Ff = 4: (Normal Reaction) = MR 5.66= 4x 875.14 u= 5.66 875.14 lx = 0.00646 or/ uk = 0.0065 Hence coefficient of kinetic friction is 0.0065
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