A small block of mass 'm' is projected horizontally with speed 'u' where friction coefficient between block and place is given by `mu = cx` where 'x' is displacement of the block on plane.
Find the displacement covered by the block before it comes to rest

Answer

To solve the problem of a small block of mass 'm' projected horizontally with speed 'u' on a plane where the friction coefficient is given by \(\mu = cx\), we need to find the displacement covered by the block before it comes to rest. Here’s a step-by-step solution:

### Step 1: Identify the forces acting on the block
The block experiences a frictional force due to the coefficient of friction, which is given as \(\mu = cx\). The normal force \(N\) acting on the block is equal to its weight, which is \(mg\). Therefore, the frictional force \(F\) can be expressed as:
\[ F = \mu N = \mu mg = cx \cdot mg \]

### Step 2: Write the equation of motion
The frictional force will cause the block to decelerate. The acceleration (or retardation) \(a\) of the block can be expressed as:
\[ a = -\frac{F}{m} = -\frac{cx \cdot mg}{m} = -cgx \]
Here, the negative sign indicates that the acceleration is in the opposite direction of the motion.

### Step 3: Relate acceleration to velocity
Using the relationship between acceleration and velocity, we have:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot v \]
Thus, we can write:
\[ \frac{dv}{dx} \cdot v = -cgx \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ v \, dv = -cgx \, dx \]

### Step 5: Integrate both sides
Integrate both sides with respect to their respective variables. The left side will be integrated from initial velocity \(u\) to final velocity \(0\), and the right side will be integrated from initial displacement \(0\) to final displacement \(x\):
\[ \int_{u}^{0} v \, dv = -cg \int_{0}^{x} x \, dx \]

### Step 6: Solve the integrals
The left side integral becomes:
\[ \left[\frac{v^2}{2}\right]_{u}^{0} = 0 - \frac{u^2}{2} = -\frac{u^2}{2} \]
The right side integral becomes:
\[ -cg \left[\frac{x^2}{2}\right]_{0}^{x} = -cg \cdot \frac{x^2}{2} \]

### Step 7: Set the integrals equal to each other
Setting the two results equal gives:
\[ -\frac{u^2}{2} = -cg \cdot \frac{x^2}{2} \]

### Step 8: Simplify and solve for \(x\)
Removing the negative signs and simplifying gives:
\[ \frac{u^2}{2} = cg \cdot \frac{x^2}{2} \]
\[ u^2 = cgx^2 \]
\[ x^2 = \frac{u^2}{cg} \]
\[ x = \frac{u}{\sqrt{cg}} \]

### Final Answer
Thus, the displacement covered by the block before it comes to rest is:
\[ x = \frac{u}{\sqrt{cg}} \]