Question

Example 6.6 A block of mass m = 1 kg,
moving on a horizontal surface with speed
v = 2 m s' enters a rough patch ranging
from x=0.10 m to x=2.01 m. The retarding
force F on the block in this range is inversely
proportional to x over this range,
-k
F = - for 0.1<x< 2.01 m
X
= 0 for x < 0.1m and x > 2.01 m
where l = 0.5 J. What is the final kinetic
energy and speed y, of the block as it
crosses this patch ?​

Answer 1

The initial kinetic energy,

$\sf{\longrightarrow K_i=\dfrac{1}{2}\,m(v_i)^2}$

$\sf{\longrightarrow K_i=\dfrac{1}{2}\cdot1 (2)^2}$

$\sf{\longrightarrow K_i=2\ J}$

By work - energy theorem, the work done on the block when it moves along the patch is equal to change in kinetic energy.

$\sf{\longrightarrow K_f-K_i=W}$

$\displaystyle\sf{\longrightarrow K_f-K_i=\int\limits_{0.10}^{2.01}F_r\ dx}$

$\displaystyle\sf{\longrightarrow K_f-K_i=-k\int\limits_{0.10}^{2.01}\dfrac{1}{x}\ dx}$

$\displaystyle\sf{\longrightarrow K_f-K_i=-k\Big[\log x\Big]_{0.10}^{2.01}}$

$\displaystyle\sf{\longrightarrow K_f-2=-0.5\cdot\log\left(\dfrac{2.01}{0.10}\right)}$

$\displaystyle\sf{\longrightarrow K_f-2=-0.5\log(20.1)}$

$\displaystyle\sf{\longrightarrow K_f= 2-0.5\times3}$

$\displaystyle\sf{\longrightarrow\underline{\underline{K_f=0.5\ J }}}$

Answer 2

$\huge{\underline{\bold{\green{ANSWER}}}}$

The initial kinetic energy,

$\sf{\longrightarrow K_i=\dfrac{1}{2}\,m(v_i)^2}$

$\sf{\longrightarrow K_i=\dfrac{1}{2}\cdot1 (2)^2}$

$\sf{\longrightarrow K_i=2\ J}$

By work - energy theorem, the work done on the block when it moves along the patch is equal to change in kinetic energy.

$\sf{\longrightarrow K_f-K_i=W}$

$\displaystyle\sf{\longrightarrow K_f-K_i=\int\limits_{0.10}^{2.01}F_r\ dx}$

$\displaystyle\sf{\longrightarrow K_f-K_i=-k\int\limits_{0.10}^{2.01}\dfrac{1}{x}\ dx}$

$\displaystyle\sf{\longrightarrow K_f-K_i=-k\Big[\log x\Big]_{0.10}^{2.01}}$

$\displaystyle\sf{\longrightarrow K_f-2=-0.5\cdot\log\left(\dfrac{2.01}{0.10}\right)}$

$\displaystyle\sf{\longrightarrow K_f-2=-0.5\log(20.1)}$

$\displaystyle\sf{\longrightarrow K_f= 2-0.5\times3}$

$\displaystyle\sf{\longrightarrow\underline{\underline{K_f=0.5\ J }}}$