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A 16.0 kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 25.0 N. Starting from rest, the sled attains a speed of 1.00 m/s in 8.00 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. You Answered

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Answer:


\mu_k = 0.15

Step-by-step explanation:

according to the kinematic equation


v^(2) - u^(2) = 2aS

Where

u is initial velocity = 0 m/s

a = acceleration

S is distance = 8.00 m

final velocity = 1.0 m/s


a = \frac {v^(2)}{2S}


a = \frac {1{2}}{2*8.6}

a = 0.058 m/s^2

from newton second law

Net force = ma


f_(net) = ma

F - f = ma

2
5 - \mu_kN = ma


25 - \mu_kmg = ma


\frac {25 - ma}{mg} =\mu_k


\frac {25 - 16*0.058}{16*9.81} = 0.15


\mu_k = 0.15

User Madhur Yadav
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