A 0.55 kg block slides across a tabletop with initial velocity of 3 m/s and it comes to rest in a distance of 0.53 m. Find the average friction force that ed its motion.

Answer 1

Wb = mg = 0.55kg * 9.8N/kg = 5.39N.

Fb = (5.39N.,0 deg.).

Fp = Fh = 5.39sin(0) = 0 = Force parallel to plane = Hor. comp.

Ff = Force due to friction.

a = (Vf^2 - V9^2) / 2d,
a = (0 - (3^2)) / 1.06 = -8.49m/s^2.

Fn = ma = 0.55 * (-8.49) = -4.67N. = Net force.

Fn = Fp - Ff = -4.67,
0 - Ff = -4.67,
Ff = 4.67N = Force due to friction.