4. A 0.422 kg object is sliding down a surface that’s inclined 12.4 with respect to the

Question

1.    A 0.422 kg object is sliding down a surface that’s inclined 12.4° with respect to the horizontal. While the object is sliding, it has a constant nonzero acceleration. Find the coefficient of kinetic friction for the surface if the object, starting from rest at the top of the surface, has a velocity of 1.33 m/s after sliding 0.566 m.

Answer 1

Since the motion along the plane acts downward on the plane we first determine the value of constant acceleration using the equations of motion. We have an initial and final velocity (u,v) and the distance covered (s) along the plane thus we use the equation

v² = u² + 2as

(1.33)² = 0² +2a(0.556)

1.7689 = 1.112a

a = 1.59

For an object sliding down the plane, the component of gravity which pulls it down is given by mgsinθ and the frictional force is μmgcosθ since the frictional force depends on the normal component of the normal force. These are the only forces acting on the object so the acceleration of the object acts downward along the plane.

The Force acting on the object is F = ma = mg sinθ - μmgcosθ

Factoring m from this equation we have

a = g sinθ - μgcosθ and substituting the given values

1.59 = 9.8 sin(12.4°) -9.8μ cos(12.4°)

1.59 = 2.1044 - 9.5713μ

-0.5144 = -9.5713μ = 0.0537