Two bloks of masses 20 kg and 50 kg are lying on a horizontal floor (coefficient of friction = 0.5).Initially string is stretched andblocksare at rest : Now two forces 300N and 150N is applied on two blocks as shown in figure. The aeleration of 20kg block is 2.5K m/s2 . Find the value of K

Question

Devansh Maheshwari · Answer

Let the tension in the rope be `T` and accleration of masses be `a`. The frictional force acting on 50kg block is 250N. And that on 20 kg block is 100N. Therefore we can say that, 300 - (250 + T) = 50a, and150 + T - 100 = 20a from here we get a = 100/70 or 1.4285 = 2.5 k , so k = 4/7

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Two bloks of masses 20 kg and 50 kg are lying on a horizontal floor (coefficient of friction = 0.5).Initially string is stretched andblocksare at rest : Now two forces 300N and 150N is applied on two blocks as shown in figure. The aeleration of 20kg block is 2.5K m/s2 . Find the value of K

Two bloks of masses 20 kg and 50 kg are lying on a horizontal floor (coefficient of friction = 0.5).Initially string is stretched andblocksare at rest : Now two forces 300N and 150N is applied on two blocks as shown in figure. The aeleration of 20kg block is 2.5K m/s2 . Find the value of K

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Two bloks of masses 20 kg and 50 kg are lying on a horizontal floor (coefficient of friction = 0.5).Initially string is stretched andblocksare at rest : Now two forces 300N and 150N is applied on two blocks as shown in figure. The aeleration of 20kg block is 2.5K m/s2 . Find the value of K

Two bloks of masses 20 kg and 50 kg are lying on a horizontal floor (coefficient of friction = 0.5).Initially string is stretched andblocksare at rest : Now two forces 300N and 150N is applied on two blocks as shown in figure. The aeleration of 20kg block is 2.5K m/s2 . Find the value of K

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