# A 29.0kg box is released on a 28 degree incline and accelerates down the incline at 0.24m/s^2. a)...

## Question:

A 29.0kg box is released on a 28 degree incline and accelerates down the incline at 0.24m/s{eq}^2 {/eq}.

a) Find the friction force impeding its motion. Express your answer to two significant figures and in Newtons.

b) Determine the coefficient of kinetic friction. Express your answer using two significant figures.

## Newton's law of motion

We have to use the concept of Kinematics of motion and newtons law. With the help of newtons law we will try to write the net force equation for box. With the help of net force equation we can figure out the friction force which later on gives us the value of coefficient of friction.

Given:

Mass of the box(m) = 29 kg

Acceleration of the block (a) = 0.24 m/s^{2}

From the newtons law we can write for the downward moving block

{eq}\displaystyle mg\ Sin\theta -f = ma {/eq}

Where f is friction force which would be

{eq}\displaystyle f = \mu*mg\ Cos\theta = \mu*(29*9.81)\ Cos28 = 251.19\mu {/eq}

Now putting this in the above equation , we get

{eq}\displaystyle (29*9.81*Sin28) - (251.19\mu ) = (29*0.24) \\ \mu = 0.504 {/eq}