Question

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

Answer 1

`\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`