Question

A block of mass 4.4 kg slides 18 m from rest down an inclined plane making an angle of 22 ^o with the horizontal. If the block takes 10 s to slide down the plane, what is the retarding force due to friction?

Answer 1

use,
s = u*t + 0.5*a*t^2
18 = 0 + 0.5*a*(10)^2
18 = 50*a
a = 0.36 m/s^2

let friction force be f
a = g*sin(20) - f/m
0.36 = 9.8*0.342 - f/4.4
0.36 = 3.35 - f/4.4
f/4.4 = 2.99
f = 13.156 N
= 13.2 N

Answer: 13.2 N