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(A) Given: Mass $m = 1000 \; kg$,speed $v = 18.0 \; km/h = 18.0 	imes (5/18) \; m/s = 5 \; m/s$,coefficient of friction $\mu = 0.5$,spring constant $

Vedclass · Accepted Answer

$1.35$

Vedclass · Answer

(A) Given: Mass $m = 1000 \; kg$,speed $v = 18.0 \; km/h = 18.0 	imes (5/18) \; m/s = 5 \; m/s$,coefficient of friction $\mu = 0.5$,spring constant $k = 6.25 	imes 10^{3} \; N/m$,and $g = 10.0 \; m/s^2$.In the presence of friction,the work-energy theorem states that the change in kinetic energy is equal to the total work done by all forces (spring force and frictional force).$\Delta K = W_{spring} + W_{friction}$$0 - \frac{1}{2} m v^2 = -\frac{1}{2} k x_m^2 - \mu m g x_m$Rearranging gives the quadratic equation: $\frac{1}{2} k x_m^2 + \mu m g x_m - \frac{1}{2} m v^2 = 0$$k x_m^2 + 2 \mu m g x_m - m v^2 = 0$Substituting the values: $(6.25 	imes 10^3) x_m^2 + 2(0.5)(1000)(10) x_m - (1000)(5^2) = 0$$6250 x_m^2 + 10000 x_m - 25000 = 0$Dividing by $1250$: $5 x_m^2 + 8 x_m - 20 = 0$Using the quadratic formula $x_m = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:$x_m = \frac{-8 + \sqrt{64 - 4(5)(-20)}}{2(5)} = \frac{-8 + \sqrt{64 + 400}}{10} = \frac{-8 + \sqrt{464}}{10} \approx \frac{-8 + 21.54}{10} = 1.354 \; m$.Thus,the maximum compression is approximately $1.35 \; m$.

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