A block of mass 5kg slides down a surface inclined at 30 degrees to the horizontal. The coefficent of

kinetic friction is 0.3. A string attached to the block is wrapped around a flywheel on a fixed axis at O.

The flywheel has a mass 30kg and a moment of inertia 0.25kg·m2 with respect to the axis of rotation. The

string pulls without slipping at a perpendicular distance of 0.4m from the axis. (a) What is the acceleration

of the block down the plant? (b) What is the tension in the string?

(a). The acceleration of the block is 1.793m/s.

(b). The string has a tension of 2.8N.

Explanation:

The forces acting on the block are the tension in the string, the frictional force , and the gravitational force , together they induce acceleration in the block; therefore,

.

Now, the frictional force is

and the tension in the string is what causes the angular acceleration in the flywheel; therefore,

where is the distance from the axis of the flywheel and is its moment of inertia.

(a).

Since the linear acceleration of the flywheel is , equation (3) gives

which when solved for gives

putting this and equation (2) into equation (1) gives

solving for we get:

Putting in , , ,, , and we get:

Thus, the acceleration of the block is 1.793m/s.

(b).

The tension in the string is given by the equation (4)

putting in  , , and we get:

Thus, the string has a tension of 2.8N.

it has to be b, about 97 percent.