A body of mass 25kg, moving at 3m/s on a rough floor horizontal floor is brought to reset after sliding through a distance of2.5m on the floor. Calculate the coefficient of sliding friction(g=10.0m/s2
Kinetic energy at start = (1/2) m v^2
Work done by friction is the same to stop it
That is Fn * 2.5 meters
so
2.5 Fn = (1/2) m (3)^2 (note I am leaving m because I think it will cancel out in the end)
Fn = m * (9/5)
but Fn = mu m g = 10 mu m
so in the end
10 mu m = (9/5) m (sure enough, m does not matter :)
mu = 9/50 = 18/100 = 0.18
initial K.E. ... 1/2 m v^2 = 1/2 * 25 * 3^2
... this equals the work done (by friction) to stop the body
work = f d = 2.5 f
plug in the K.E. for work and solve for the force (f)
the coefficient is ... f / (m g)