If a 60.0 kg person slides down an incline angled 25˚ below the horizontal and friction IS present with a coefficient of kinetic friction equal to k = 0.450. What is the person's acceleration as they slide down the slide.

Answer 1

Wp = m*g = 60kg * 9.8N/kg = 588 N. = Wt. of the person.

Fp = 588*sin25 = 248.5 N. = Force parallel to plane.

Fv = 588*cos25 = 533 N. = Force perpendicular to plane

Fk = 0.450 * 533 = 240 N. = Force of kinetic friction.

Fn = m*a
a = Fn/m = (Fp-Fk)/m
a = (248.5-240)/60 = 0.142 m/s^2.