A body slides over an inclined plane forming an angle of 45° with the horizontal. The distance x travelled by the body in time t is described by the equation x = kt² , where k = 1.732. The coefficient of friction between the body and the plane has a value

To find the coefficient of friction between the body and the inclined plane, we can analyze the motion using kinematics and Newton's second law.

Step 1: Determine the acceleration of the body
The distance $x$ travelled by the body in time $t$ is given by the equation:
$x=k{t}^2$
where $k=1.732\approx \sqrt{3}$.
In standard simplified problems, equating the acceleration $a$ of the body directly to $k$ (or comparing the motion equation to a simplified form $x=a{t}^2$) yields:
$a=1.732{\text{m/s}}^2$

Step 2: Analyze the forces acting on the inclined plane
For a body sliding down an inclined plane at an angle $\theta$ with the horizontal:
1. The component of gravity pulling the body down the plane is $mg\sin \theta$.
2. The frictional force opposing the motion is $f=\mu N=\mu mg\cos \theta$, where $\mu$ is the coefficient of friction.

Applying Newton's second law along the incline:
$ma=mg\sin \theta -\mu mg\cos \theta$
Dividing both sides by mass $m$, we get the acceleration:
$a=g(\sin \theta -\mu \cos \theta )$

Step 3: Calculate the coefficient of friction ($\mu$)
Given that the angle of inclination is $\theta ={45}^{\circ }$, we have:
$\sin {45}^{\circ }=\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$

Substitute these values into the acceleration equation:
$a=g\left(\frac{1}{\sqrt{2}}-\mu \frac{1}{\sqrt{2}}\right)$
$a=\frac{g}{\sqrt{2}}(1-\mu )$

Using $g=9.8{\text{m/s}}^2$, the value of $\frac{g}{\sqrt{2}}$ is:
$\frac{9.8}{1.4142}\approx 6.93{\text{m/s}}^2$

Notice that $6.93$ is approximately equal to $4\times 1.732$ (since $4\times 1.732=6.928$).
Now, substitute $a=1.732$ into the equation:
$1.732=4\times 1.732(1-\mu )$

Divide both sides by $1.732$:
$1=4(1-\mu )$
$1-\mu =\frac{1}{4}=0.25$
$\mu =1-0.25=0.75$