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(C) The body is placed on a moving belt. Initially, the body is at rest relative to the ground, but it has a velocity relative to the belt. The kineti

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$1$

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(C) The body is placed on a moving belt. Initially, the body is at rest relative to the ground, but it has a velocity relative to the belt. The kinetic friction force $f_k$ acts on the body to oppose this relative motion.$f_k = \mu_k N = \mu_k mg$Using Newton's second law, the deceleration $a$ of the body relative to the belt is:$ma = \mu_k mg$$a = \mu_k g = 0.2 	imes 10 = 2 \,m \,s^{-2}$The initial velocity of the body relative to the belt is $u = 2 \,m \,s^{-1}$. The body comes to rest relative to the belt when its final velocity $v = 0$.Using the equation of motion $v^2 = u^2 - 2as$:$0^2 = (2)^2 - 2(2)s$$4 = 4s$$s = 1 \,m$Thus, the distance travelled by the body before coming to rest relative to the belt is $1 \,m$.

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