Question

The block of a mass 10.2 kg is sliding at an initial velocity of 3.40 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.153.

Answer 1

m = mass = 10.2 kg

vo = initial velocity = 3.40 m/s

u = coefficient of kinetic friction = 0.153

g= gravity = 9.8m/s^2

a)

Fr = force of kinetic friction = u m g

Fr = 0.153 x 10.2 x 9.8 = -15.30N

b) Block's acceleration

Newton's second law of motion:

F = m*a

a = F/m = -15.30 / 10.2 = -1.5 m/s^2

c) USe the third equation of motion:

2as = vf^2 - vo^2

Where:

Vf= final velocity= 0 m/s

s = displacement

2 * -1.5 * s = 0^2 - 3.40^2

-3s = -11.56

s= -11.56/-3

s= 3.85 m