Subtract the final kinetic energy from the work done pushing the book, 25*0.88 joules
The difference will be work done against friction, which is Ff*X
Ff is the friction force, M*g*Muk
Solve for Muk
Can't figure this one out.
"In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.88 m by the horizontal 25 N force from the broom and then has a speed of 1.63 m/s, what is the coefficient of kinetic friction between the book and floor?"
I know these values, but I just can't find a formula that uses velocity and distance to plug them into.
m = 3.5 kg
g = 9.8 m/s/s
d = 0.88 m
F = 25N
Fn = 34.3N (Fn = mg)
v = 1.63 m/s
a = 7.14 m/s/s (a = F/m)
Muk = ??
I'm not sure if I even need all of those values...can anyone help me out with this?
3 answers
I used my last attempt at the problem using this info, but was unable to figure out the answer. Thank you for responding though.
W = 22 J = total work done
Final KE = (1/2)MV^2 = 4.65 J
Friction work done = 22-4.65 = 17.35 J
Friction force = 17.35/X = 19.7 N
Muk = 19.7/(M*g) = 0.575
(Muk is a dimensionless ratio)
Final KE = (1/2)MV^2 = 4.65 J
Friction work done = 22-4.65 = 17.35 J
Friction force = 17.35/X = 19.7 N
Muk = 19.7/(M*g) = 0.575
(Muk is a dimensionless ratio)
1 answer