Question

A block is placed on a plane inclined at 37.0° to the horizontal. The coefficient of kinetic friction is 0.40. If the block is released, what is its acceleration? include a diagram.

Answer 1

2.77 m/s^2

Step-by-step explanation:

For simplicity, we will say that the x-axis is parallel to the inclined plane and the y-axis is perpendicular to the inclined plane. The force that the surface applies on the block N will be perpendicular to the plane. The force of friction can be expressed like this:

`Fr = u*N`

Where u is the coefficient of kinetic friction.

The x-component and y-component of the weight force will be:

`W_y = W*cos(37)\\W_x = W*sin(37)`

In the y-axis, the acceleration of the block is 0. Then:

`N - W_y = 0\\N = W_y = W*cos(37) = m*g*cos(37)`

In the X-axis, the summation of forces would be:

`W_x - Fr = m*a\\m*g*sin(37) - u*N = m*g*sin(37) - u*m*g*cos(37) = ma\\a = (mg(sin(37)-u*cos(37)))/(m) = 9.81m/s^2*(sin(37)-0.4*cos(37))=2.77m/s^2`