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

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