° with the horizontal. The sphere has a diameter of 10 cm and a mass of 1 kg. What is the acceleration of the sphere down the incline?

We can start by analyzing the forces acting on the sphere. There are two forces: the gravitational force (mg) and the normal force (N) from the incline. Since the sphere is rolling without slipping, there is also a frictional force (f) that opposes the motion and acts at the point of contact between the sphere and the incline.

The gravitational force can be resolved into two components: mg sin(60°) down the incline and mg cos(60°) perpendicular to the incline. The normal force is equal in magnitude and opposite in direction to the perpendicular component of the gravitational force, so N = mg cos(60°).

The frictional force f is given by the equation f = μN, where μ is the coefficient of friction between the sphere and the incline. Since the sphere is rolling without slipping, we can relate the linear acceleration a of the sphere to its angular acceleration α and radius r by the equation a = αr. We can also relate the angular acceleration α to the torque τ on the sphere by the equation τ = Iα, where I is the moment of inertia of the sphere.

For a solid sphere, the moment of inertia is given by I = (2/5)mr^2, where m is the mass of the sphere and r is the radius. Plugging in the given values, we get I = (2/5)(1 kg)(0.05 m)^2 = 0.0005 kg·m².

The torque on the sphere is due to the frictional force f, which acts at a distance of r/2 from the center of the sphere. Thus, we have τ = f(r/2), or f = 2τ/r. Substituting for f and τ, we get μN = 2Iα/r². Solving for α, we get α = (μN r²)/(2I).

Now we can use the relationship between linear and angular acceleration to find the acceleration of the sphere down the incline. Since the sphere is rolling without slipping, we have a = αr, or a = (μN r)/5m. Substituting for N and simplifying, we get a = (μg sin(60°))/5.

Finally, we can plug in the given values to get the numerical answer. Using a coefficient of friction of μ = 0.2, we get a = (0.2)(9.8 m/s²)(sin(60°))/5 = 0.84 m/s². Therefore, the acceleration of the sphere down the incline is 0.84 m/s².