Post a New Question

Physics

posted by on .

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.

  • Physics - ,

    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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question