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A block having mass M slides down an inclined plane. The force of friction between the block and the inclined plane is f, the block's weight is Mg, and the normal force is N. (a) Draw a free body force diagram showing the forces acting on the block. (b) Write down all relevant Newton's equations for a given situation.

A 9.50 cm high elevation block is placed under one end of a 1.00 m long track. Sketech blem, and find the angle of elevation of the track.

User Zhiyan
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Final answer:

The free body diagram consists of the gravitational force, normal force, and frictional force. The angle of elevation can be found using sine, and the coefficient of friction is a ratio of the frictional force to the normal force.

Step-by-step explanation:

When a block slides down an inclined plane, there are several forces acting on it. A free body diagram should show the force of gravity (Mg), acting downward, the normal force (N), acting perpendicular to the surface of the incline, and the force of friction (f), acting opposite to the direction of motion. In this case, the gravitational force is decomposed into two components: one perpendicular to the plane (Mg cos(\theta)), which is balanced by the normal force, and one parallel to the plane (Mg sin(\theta)), which contributes to the block sliding down the plane when not counteracted by friction.

To find the angle of elevation of a track, we can use trigonometry. The elevation (height) of the block and the length of the track form a right triangle. The angle \( \theta \) of the incline can be calculated using the sine function: \( \sin(\theta) = \dfrac{opposite}{hypotenuse} = \dfrac{9.50\,\text{cm}}{1.00\,\text{m}} \). Convert the height into meters to compute the angle.

The coefficient of friction can be calculated by dividing the frictional force by the normal force, which comes from the perpendicular component of gravity. The coefficient represents the ratio of the force of friction to the normal force and can be found using the equation \( \mu = \frac{f}{Mg cos(\theta)} \).

User Srikanth P
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