Ask Question
10 January, 01:11

A 6.0-kg box slides down an inclined plane that makes an angle of 39o with the horizontal. If the coefficient of kinetic friction is 0.40, at what rate does the box accelerate down the slope?3.1m/s23.4m/s23.7m/s24.1m/s2

+1
Answers (1)
  1. 10 January, 03:29
    0
    A. 3.1m/s²

    Explanation:

    Since the body is sliding down the plane, the frictional force (Ff) being a force of opposition will be acting upwards.

    The forces acting on the body in the vertical direction will be the weight (W) and the normal reaction (R) acting in the opposite direction.

    Taking the sum of the forces along the plane,

    Fm - Ff = mass * acceleration (newton's law)

    Since Fm = Wsin (theta)

    Ff = nR where n is the coefficient of friction we will have;

    Wsin (theta) - nR = ma ... (1)

    W = mg = 6*10 = 60N

    n = 0.40

    R = Wcos (theta) {resolving weight to the vertical)

    R = 60cos39°

    R = 46.6N

    Substituting this values into eqn 1 to the get the acceleration,

    60sin39° - 0.4 (46.6) = 6a

    37.8 - 18.64 = 6a

    19.16 = 6a

    a = 19.16/6

    a = 3.19m/s²

    The acceleration of the body down the plane will be 3.1m/s²
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 6.0-kg box slides down an inclined plane that makes an angle of 39o with the horizontal. If the coefficient of kinetic friction is 0.40, ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers