A block of mass 10 , kg,moving in x direction | vedclass

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(A) Given: Mass $m = 10 \, kg$,initial velocity $v_i = 10 \, m/s$.Initial kinetic energy $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} 	imes 10 	imes (10

Vedclass · Accepted Answer

$475$

Vedclass · Answer

(A) Given: Mass $m = 10 \, kg$,initial velocity $v_i = 10 \, m/s$.Initial kinetic energy $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} 	imes 10 	imes (10)^2 = 500 \, J$.The retarding force is $F = -0.1 \, x \, N$.Work done by the retarding force $W = \int_{20}^{30} (-0.1 \, x) \, dx$.$W = -0.1 \left[ \frac{x^2}{2} ight]_{20}^{30} = -0.1 	imes \frac{1}{2} (30^2 - 20^2) = -0.05 	imes (900 - 400) = -0.05 	imes 500 = -25 \, J$.According to the work-energy theorem,$W = K_f - K_i$.$K_f = W + K_i = -25 \, J + 500 \, J = 475 \, J$.

$A$ block of mass $10 \, kg$,moving in $x$ direction with a constant speed of $10 \, m/s$,is subjected to a retarding force $F = 0.1 \, x \, N$ during its travel from $x = 20 \, m$ to $30 \, m$. Its final $KE$ will be ............... $J$.

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