Compute the acceleration of the block sliding down a 30 deg inclined plane if the coefficient of kinetic friction is 0.20.
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To compute the acceleration of the block sliding down an inclined plane, we need to consider the forces acting on the block. The key forces involved are the gravitational force (mg) acting downward and the force of kinetic friction (fk) acting in the opposite direction of motion.
First, let's resolve the gravitational force into its components. The component parallel to the incline is mg*sin(30°) and the component perpendicular to the incline is mg*cos(30°).
The force of kinetic friction (fk) can be computed using the equation fk = μk * N, where μk is the coefficient of kinetic friction and N is the normal force. The normal force (N) is equal to mg*cos(30°) (since the block is on an inclined plane).
Now, let's find the net force acting on the block in the direction of motion. This can be calculated using the equation: net force = force_parallel - force_friction.
So, net force = mg*sin(30°) - fk.
Since the block is moving down the inclined plane, the net force is equal to the mass (m) times the acceleration (a) of the block.
Setting these equations equal, we have: mg*sin(30°) - fk = ma.
Finally, we can rearrange the equation to solve for acceleration (a): a = (mg*sin(30°) - fk) / m.
Plugging in the given values, where the coefficient of kinetic friction (μk) is 0.20 and the angle (θ) is 30 degrees, we have:
a = (m * g * sin(30°) - μk * m * g * cos(30°)) / m.
Simplifying the equation gives:
a = g * (sin(30°) - μk * cos(30°)).
Now, we can substitute the values of g (acceleration due to gravity) and solve for a:
a = 9.8 m/s^2 * (sin(30°) - 0.20 * cos(30°)).
Calculating the expression gives:
a ≈ 9.8 m/s^2 * (0.5 - 0.20 * 0.866).
Simplifying further, we have:
a ≈ 9.8 m/s^2 * (0.5 - 0.1732).
Evaluating the expression:
a ≈ 9.8 m/s^2 * (0.3268).
Therefore, the acceleration of the block sliding down the 30° inclined plane with a coefficient of kinetic friction of 0.20 is approximately 3.19 m/s^2.