posted by janell .
2. a body of mass 20kg is released from rest t the top of a rough slope which is inclined at 30 degrees to the horizontal.six seconds later the body has a velocity of 21 m.s down the slope. find the resistance to motion in newtons experienced by the body?
Fb=mg = 20kg * 9.8N/kg = 196N @ 30deg.
Fp = 196*sin30 = 98N = Force parallel
a=(Vf-Vo) / t = (21-0) / 6 = 3.5m/s^2.
Fn = Fp-Ff = ma,
98 - Ff = 20 * 3.5,
Ff = 28N = Force of friction.