Two blocks A and B of masses 5 kg and 10 kg | vedclass

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(A) Given: $m_A = 5 \ kg$,$m_B = 10 \ kg$,$\mu = 0.2$,$F = 100 \ N$,$	heta = 30^{\circ}$,$g = 10 \ m/s^2$.Horizontal component of force $F_x = F \cos

Vedclass · Accepted Answer

$2.6$

Vedclass · Answer

(A) Given: $m_A = 5 \ kg$,$m_B = 10 \ kg$,$\mu = 0.2$,$F = 100 \ N$,$	heta = 30^{\circ}$,$g = 10 \ m/s^2$.Horizontal component of force $F_x = F \cos 30^{\circ} = 100 	imes \frac{\sqrt{3}}{2} = 50\sqrt{3} \approx 86.6 \ N$.Vertical component of force $F_y = F \sin 30^{\circ} = 100 	imes 0.5 = 50 \ N$.Normal force on $A$ from ground $N_A = m_A g + F_y = 50 + 50 = 100 \ N$.Friction on $A$ from ground $f_A = \mu N_A = 0.2 	imes 100 = 20 \ N$.Normal force between $A$ and $B$ is $N_{AB} = F_x = 86.6 \ N$.Friction between $A$ and $B$ is $f_{AB} = \mu N_{AB} = 0.2 	imes 86.6 = 17.32 \ N$.Equation for $A$: $F_x - f_A - f_{AB} = m_A a \Rightarrow 86.6 - 20 - 17.32 = 5a \Rightarrow 49.28 = 5a \Rightarrow a = 9.856 \ m/s^2$.Equation for $B$: $f_{AB} - f_B = m_B a$,where $f_B = \mu (m_B g) = 0.2 	imes 100 = 20 \ N$.$17.32 - 20 = 10a$. Since $f_{AB} < f_B$,block $B$ does not move relative to the wall. Thus,acceleration of $B$ is $0 \ m/s^2$. Given the options,there is a discrepancy in the problem statement parameters. Assuming the intended question asks for the acceleration of the system if they move together: $a = \frac{F_x - f_{total}}{m_A + m_B} = \frac{86.6 - (20 + 20)}{15} = \frac{46.6}{15} \approx 3.1 \ m/s^2$. Given the options provided,the closest value is $2.6 \ m/s^2$.

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