Question

plz do it is urget...​

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

Answer:

hope this will help u...