Question

if kinetic energy K depends on velocity V acceleration a and density P of an object find the expression for K​

Answer 1

To express Kinetic Energy K in terms of velocity, acceleration and density of an object:

Let Kinetic energy be dependent on these physical quantities as follows:

$\therefore \: K \: \propto \: {v}^{x} \times {a}^{y} \times { \rho}^{z}$

$= > \: \bigg \{ M {L}^{2} {T}^{ - 2} \bigg \}\: \propto \: { \bigg \{L{T}^{ - 1} \bigg \}}^{x} \times { \bigg \{ L{T}^{ - 2} \bigg \}}^{y} \times { \bigg \{M{L}^{ - 3} \bigg \}}^{z}$

$= > \: \bigg \{ M {L}^{2} {T}^{ - 2} \bigg \}\: \propto \: { \bigg \{ M\bigg \}}^{z} \times { \bigg \{ L\bigg \}}^{x + y - 3z} \times { \bigg \{ T\bigg \}}^{ - x - 2y}$

Comparing the powers on both sides , we get the following Equations ;

$1) \: z = 1 \\ 2) \: x + y - 3z = 2 \\ 3) \: - x - 2y = - 2$

Solving these Equations , we get ;

$1) \: z = 1 \\ 2) \: y = - 3 \\ 3)x = 8$

So putting the values ;

$\therefore \: K \: \propto \: {v}^{8} \times {a}^{ - 3} \times { \rho}^{1}$

Putting a constant "c" ;

$= > \: K \: = c \bigg( \: {v}^{8} \times {a}^{ - 3} \times { \rho}^{1} \bigg)$

So, final answer is:

$\boxed{ \sf{ \bold{ \red{\: K \: = c \bigg( \: {v}^{8} \times {a}^{ - 3} \times { \rho}^{1} \bigg)}}}}$