A basket and its contents have mass M. A monk | vedclass

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Question

(A) Let $v_m$ be the velocity of the monkey and $v_b$ be the velocity of the basket relative to the ground. Since the rope is inextensible,the speed o

Vedclass · Accepted Answer

$v_{basket} = 4/3 \, ft/s$

Vedclass · Answer

(A) Let $v_m$ be the velocity of the monkey and $v_b$ be the velocity of the basket relative to the ground. Since the rope is inextensible,the speed of the monkey relative to the rope is $v_{m/r} = v_m - (-v_b) = v_m + v_b$. Given $v_{m/r} = 2 \, ft/s$,we have $v_m + v_b = 2$. Since the system is initially at rest and the monkey pulls the rope,the center of mass of the system (monkey + basket) remains stationary if no external force acts,but here gravity acts. However,the internal tension forces cancel out. Using the conservation of momentum for the system in the vertical direction: $m_m v_m + m_b v_b = 0$. Substituting $m_m = 2M$ and $m_b = M$: $(2M)v_m + (M)v_b = 0 \implies 2v_m + v_b = 0 \implies v_m = -v_b/2$. Substituting this into the relative velocity equation: $-v_b/2 + v_b = 2 \implies v_b/2 = 2 \implies v_b = 4 \, ft/s$. Wait,re-evaluating: The monkey pulls the rope. The tension $T$ acts on both. For the monkey: $T - 2Mg = 2Ma_m$. For the basket: $T - Mg = Ma_b$. Since the rope length is constant,$a_m = -a_b$. $T - 2Mg = 2M(-a_b) \implies T = 2Mg - 2Ma_b$. Substituting into the basket equation: $(2Mg - 2Ma_b) - Mg = Ma_b \implies Mg = 3Ma_b \implies a_b = g/3 = 32/3 \, ft/s^2$. After $t = 3 \, s$,$v_b = a_b t = (32/3) 	imes 3 = 32 \, ft/s$. However,the question specifies the monkey moves at a constant speed relative to the rope. This implies the acceleration phase is instantaneous. If $v_{m/r} = 2$ is constant,then $v_m + v_b = 2$. With $2v_m + v_b = 0$,we get $v_b = 4 \, ft/s$.

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