અચળ ઝડપ 10;ms^{-1} થી x - દિશામાં 10 ;kg દળનો | vedclass

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Question

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,Here,\,m = 10\,kg,\,{v_i} = 10\,m\,{s^{ - 1}}\
Initial\,kinetic\,energy\,of\,the\,block\,is\
{K_i} = \fra

Vedclass · Accepted Answer

$475$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,Here,\,m = 10\,kg,\,{v_i} = 10\,m\,{s^{ - 1}}\
Initial\,kinetic\,energy\,of\,the\,block\,is\
{K_i} = \frac{1}{2}mv_i^2 = \frac{1}{2} 	imes \left( {10\,kg} ight) 	imes {\left( {10\,m{s^{ - 1}}} ight)^2}\
\,\,\,\,\,\, = \,500\,J\
Work\,done\,by\,retarding\,force\
W = \int\limits_{{x_1}}^{{x_2}} {{F_r}dx = \int\limits_{20}^{30} { - 0.1\,xdx =  - 0.1\left[ {\frac{{{x^2}}}{2}} ight]_{20}^{30}} } 
\end{array}$

$\begin{array}{l}
 =  - 0.1\left[ {\frac{{900 - 400}}{2}} ight] =  - 25\,J\
According\,to\,work - energy\,theorem,\
W = {K_f} - {K_i}\
{K_f} = W + {K_i} =  - 25\,J + 500\,J = 475\,J
\end{array}$

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