Question

12.5 kg wooden crate with an initial velocity of 2.5 m/s slides across a rough cement floor for 1.7 m before coming to rest. Calculate the coefficient of kinetic friction

Answer 1

x = x0 + v0*t - 1/2*a*t^2
v = v0 - a*t
m*a = m*g*uk

where x is distance, x0 is the inital x position, v is speed, v0 is the initial speed, a is the acceleration of the crate, t is time, m is the mass of the crate, g is the acceleration due to gravity, and uk is the coefficient of kinetic friction

When the box stops, v = 0:

0 = 2.5 - a * t
a = g*uk

0 = 2.5 - 9.8*t*uk
1.7 = 2.5*t - .5*9.8*uk*t^2

Solve this system of equations for t and uk

Answer 2

.19