A 40 kg child start from rest and travels 4 m down a slide that is inclined 35 degree to the horizontal. If her speed at the bottom is 2.5 m/s , what was the frictional force along the slide?
M*g = 40 * 9.8 = 392 N. = Wt. of child.
Fp = 392*sin35 = 224.8 N. = Force parallel to the incline.
V^2 = Vo^2 + 2a*d = 2.5^2.
0 + 2a*4 = 6.25, a = 0.781 m/s^2.
Fp-Fk = M*a.
Fk = Fp-M*a = 224.8-40*0.781 = 193.6 N.