mv²/2 =W(fr) = μNs= μmgs
μ=v²/2gs =5.8²/2•9.8•20 =0.086
A hockey player hits a puck with his stick, giving the puck an initial speed of 5.8m/s .
Part A
If the puck slows uniformly and comes to rest in a distance of 20m , what is the coefficient of kinetic friction between the ice and the puck?
2 answers
x = (v-u)(v+u)/2a
where x = distance, v = final speed, u = initial speed, a = acceleration
a = (v-u)(v+u)/2x = -5.8*5.8/40 = -0.841
f=ma
u(k) * f(normal) = ma
u(k) *m *g = ma
u(k) *g = a
u(k) = a/g
u(k) = -0.841 / 9.81 = 0.086
where u(k) = coefficient of kinetic friction
where x = distance, v = final speed, u = initial speed, a = acceleration
a = (v-u)(v+u)/2x = -5.8*5.8/40 = -0.841
f=ma
u(k) * f(normal) = ma
u(k) *m *g = ma
u(k) *g = a
u(k) = a/g
u(k) = -0.841 / 9.81 = 0.086
where u(k) = coefficient of kinetic friction