A frictionless cart A of mass 100 kg carries | vedclass

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(C) Let $a$ be the acceleration of the entire system in the right direction. Since the smaller carts $B$ and $C$ do not move relative to the larger ca

Vedclass · Accepted Answer

$560$

Vedclass · Answer

(C) Let $a$ be the acceleration of the entire system in the right direction. Since the smaller carts $B$ and $C$ do not move relative to the larger cart $A$,the equation of motion for the whole system is: $(m_A + m_B + m_C) a = F$ $(100 + 8 + 4) a = F$ $112 a = F \quad ...(1)$ For block $B$,the tension $T$ provides the acceleration $a$: $T = m_B a = 8a \quad ...(2)$ For block $C$,the tension $T$ balances the gravitational force: $T = m_C g = 4 	imes 10 = 40 \ N \quad ...(3)$ Equating $(2)$ and $(3)$: $8a = 40$ $a = 5 \ m/s^2$ Substituting $a$ into equation $(1)$: $F = 112 	imes 5 = 560 \ N$

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