Question

A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts to slide down the incline, traveling 3.40 m down the incline in 2.00 s. Calculate the coefficient of static friction between the block and the plank.
Motion of a block?

Calculate the kinetic coefficent of friction between the block and the plank?

Answer 1

Answer:

$\mu_s = 0.62$

$\mu_k = 0.415$

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

$x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2$

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

$F = ma\\F = 1.7m$

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

$F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415$

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

$mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62$