posted by Andrew on .
A person pushes horizontally with a force of 170 N on a 68 kg crate to move it across a level floor. The coefficient of kinetic friction is 0.22. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration?
Wc = mg = 68kg * 9.8N/kg = 666.4N. =
Weight of crate.
Fc = (666.4N,0deg.).
Fp = Fh = 666.4sin(0) = 0 Newtons. =
Force parallel to floor = Hor. cmp.
Fv = ver. = 666.4cos(0) = 666.4N. =
Force perpendicular to floor.
a. Ff = u*Fv = 0.22 * 666.4 = 147N. = Force of friction.
b. Fn = Fap - Ff = 170 - 147 = 23N.
Fn = ma,
a = Fn/m = 23 / 68 = 0.39m/s^2.
Correction: a = 23 / 68 = 0.34m/s^2.