A water channel of width b = 17 m lies between two frictionless platforms A and B. A plank (P) of length $l = 2.0 m$ rest near the platform A. A block of mass m = 25 kg sliding on the platform A lands on the plank and stays on it. Force $\vec{F}$ of water resistance on the plank is proportional to velocity of the plank $({\vec{v}}_{p r})$ relative to the water and given by the law $\vec{F} = - k {\vec{v}}_{p r}$ , where k = 15 N–s/m. Calculate the minimum speed $u_{\min}$ of the block on platform A so that plank reaches the platform B.
(Answer in m/s)

Question

A water channel of width b = 17 m lies between two frictionless platforms A and B. A plank (P) of length $l = 2.0 m$ rest near the platform A. A block of mass m = 25 kg sliding on the platform A lands on the plank and stays on it. Force $\vec{F}$ of water resistance on the plank is proportional to velocity of the plank $({\vec{v}}_{p r})$ relative to the water and given by the law $\vec{F} = - k {\vec{v}}_{p r}$ , where k = 15 N–s/m. Calculate the minimum speed $u_{\min}$ of the block on platform A so that plank reaches the platform B.
(Answer in m/s)

Accepted Answer

Since water resistance (F→=−kv→pr)  is proportional to the velocity of plank relative to the water, impulse of water resistance equals to product of the proportionality constant k and displacement of the plank relative to the water. And this impulse stops the block–plank system at the other bank, therefore equals to the initial momentum of the block.mumin=k∫v  dt⇒mumin=k(b−l)umin=k(b−l)m=9.0  m/s

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