A 35.4 kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.323. The acceleration of gravity is 9.8 m/s^2. If the crate is pushed horizontally with a force of 186.012 N, how far does it move in 5.11 s? Answer in units of m.
+37146 Another way w/out calculus
Normal force = 9.81 m/s^2 * 35.4 kg
frictional force = friction coefficient * normal force = .323 * ( 9.81 * 35.4)
net force on object = 186.012 - .323*(9.81*35.4) = 73.84 N
F = ma
73.84 = 35.4 * a a = 2.086 m/s^2
x = x0 + v0 t + 1/2 a t^2
= 0 + 0 + 1/2 (2.086)(5.11^2) = 27 .23 meters
