A force F = Kt (where t is the time in second | vedclass

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Question

(A) Let the common acceleration of the system be $a$. For the $8 \, kg$ block,the only horizontal force is the static friction $f_s$ exerted by the $2

Vedclass · Accepted Answer

$\frac{125}{240} \, m$

Vedclass · Answer

(A) Let the common acceleration of the system be $a$. For the $8 \, kg$ block,the only horizontal force is the static friction $f_s$ exerted by the $2 \, kg$ block. $f_s = m_2 a = 8a$. For the $2 \, kg$ block,the net force is $F - f_s = m_1 a$. $2t - 8a = 2a \implies 2t = 10a \implies a = \frac{t}{5}$. The blocks start slipping when the static friction reaches its maximum value: $f_{s,max} = \mu N = 0.2 	imes 2 	imes 10 = 4 \, N$. Equating $f_s = 8a = 4 \, N$,we get $a = 0.5 \, m/s^2$. Since $a = \frac{t}{5}$,we have $0.5 = \frac{t}{5} \implies t = 2.5 \, s$. Now,$a = \frac{dv}{dt} = \frac{t}{5}$. Integrating with respect to time: $v = \int_0^t \frac{t}{5} dt = \frac{t^2}{10}$. Displacement $x = \int_0^t v dt = \int_0^{2.5} \frac{t^2}{10} dt = \left[ \frac{t^3}{30} ight]_0^{2.5} = \frac{(2.5)^3}{30} = \frac{15.625}{30} = \frac{15625}{30000} = \frac{125}{240} \, m$.

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