Question

A block of mass m moving with a velocity $\mathrm{v}_{0}$ on a rough horizontal surface from $\mathrm{x}=0 .$Coefficient of friction is given by $\mu=\mu_{0}+\mathrm{ax} .$ Find the kinetic energy of the block as a function of $x$ before it comes to rest.

Answer 1

Hint: An object's kinetic energy is the energy it possesses due to its motion. It is defined as the work required to accelerate from rest to the specified velocity of a body of a given mass. Having accumulated this energy through its acceleration, until its speed increases, the body retains this kinetic energy. A body or a system's total kinetic energy is equal to the sum of kinetic energies resulting from each form of motion.
Formula used:
Work done $=\Delta \mathrm{K}$

Complete answer:
Work energy theorem Often known as the theory of work and kinetic energy, the work-energy theorem states that the total work done by the sum of all the forces acting on a particle is equal to the change in the particle's kinetic energy. The theorem of work-energy states that the net work performed by the forces on an object is equal to the change in its kinetic energy.
Friction(f) of the body is $=\mu \text{N}$
$=\mu \mathrm{N}=\mu \mathrm{mg}=\left(\mu_{0}+\alpha \mathrm{x}\right) \mathrm{mg}$
$=\mu \mathrm{N}=\mu \mathrm{mg}=\left(\mu_{0}+\alpha \mathrm{x}\right) \mathrm{mg}$
$\mathrm{W}=$ work done by the frictional force
$\Rightarrow -\int_{0}^{x}{\left( {{\mu }_{0}}+\alpha x \right)}\operatorname{mgdx}$$=-\int_{0}^{x}\left(\mu_{0}+\alpha x\right) \operatorname{mgdx}=-m g\left(\mu_{0} \cdot x+\dfrac{\alpha x^{2}}{2}\right)$
Applying work energy theorem
Work done = $\Delta \text{K}$
$\Rightarrow -\text{mg}\left( {{\mu }_{0}}\cdot \text{x}+\dfrac{\alpha {{\text{x}}^{2}}}{2} \right)$$-\mathrm{mg}\left(\mu_{0} \cdot \mathrm{x}+\dfrac{\alpha \mathrm{x}^{2}}{2}\right)=\dfrac{1}{2} \mathrm{mv}^{2}-\left(\dfrac{1}{2}\right) \mathrm{mv}_{0}^{2}$
$\therefore \dfrac{1}{2}\text{m}{{\text{v}}^{2}}=\left( \dfrac{1}{2} \right)\text{mv}_{0}^{2}-\text{mg}\left( {{\mu }_{0}}\cdot \text{x}+\dfrac{\alpha {{\text{x}}^{2}}}{2} \right)$
\[\therefore \] The kinetic energy of the block as a function of $x$ before it comes to rest is $\dfrac{1}{2} \mathrm{mv}^{2}=\left(\dfrac{1}{2}\right) \mathrm{mv}_{0}^{2}-\mathrm{mg}\left(\mu_{0} \cdot \mathrm{x}+\dfrac{\alpha \mathrm{x}^{2}}{2}\right)$

Note:
Kinetic energy is directly proportional to the mass of the object and the square of its velocity: if the mass consists of units of kilograms and the velocity of meters per second, the kinetic energy consists of units of kilograms per square second \[\text{KE=}\dfrac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\] While it is in motion, the human body also generates kinetic energy. ... For experiments, kinetic energy has often been used and then transferred to other sources of energy, which are then used to power everything.