Block of mass 5kg is moving with velocity –5 ˆk (m/s) relative to the platform as shown in figure (block is in contact with vertical surface also). Vertical and horizontal surfaces of large platform are having friction coefficient 1/2 and 1/4 respectively. The platform is being moved with constant acceleration ˆ 5i (m/s2 ) select correct alternative/alternatives : (g = 10 m/s2 ) (A) Block will stop after travelling 5/2 m relative to platform (B) Block will stop after travelling 5 m relative to platform (C) Time after which the block will stop with respect to platform is 2 sec. (D) Time after which the block will stop with respect to platform is 1 sec
Question
Block of mass 5kg is moving with velocity –5 ˆk (m/s) relative to the platform as shown in figure (block is in contact with vertical surface also). Vertical and horizontal surfaces of large platform are having friction coefficient 1/2 and 1/4 respectively. The platform is being moved with constant acceleration ˆ 5i (m/s2 ) select correct alternative/alternatives : (g = 10 m/s2 ) (A) Block will stop after travelling 5/2 m relative to platform (B) Block will stop after travelling 5 m relative to platform (C) Time after which the block will stop with respect to platform is 2 sec. (D) Time after which the block will stop with respect to platform is 1 sec
Solution
The problem involves a block moving on a platform with both horizontal and vertical friction. The block is initially moving with a velocity of -5 m/s (in the k direction) and the platform is accelerating at 5 m/s^2 (in the i direction). The friction coefficients for the vertical and horizontal surfaces are 1/2 and 1/4 respectively.
Let's solve this step by step:
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First, we need to find the frictional force acting on the block. The normal force on the block is its weight, which is mass times gravity, or 5kg * 10m/s^2 = 50N.
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The frictional force on the horizontal surface is the coefficient of friction times the normal force, or (1/4) * 50N = 12.5N.
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The frictional force on the vertical surface is (1/2) * 50N = 25N.
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The total frictional force acting on the block is the sum of these two forces, or 12.5N + 25N = 37.5N.
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This force will decelerate the block. The acceleration due to friction is the frictional force divided by the mass of the block, or 37.5N / 5kg = 7.5m/s^2.
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The block is initially moving at 5m/s, so the time it takes for the block to stop is the initial velocity divided by the acceleration, or 5m/s / 7.5m/s^2 = 0.67 seconds.
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The distance the block travels before it stops is the initial velocity times the time it takes to stop, or 5m/s * 0.67s = 3.33 meters.
So, none of the options (A, B, C, D) are correct. The block will stop after travelling approximately 3.33 meters relative to the platform, and it will stop approximately 0.67 seconds after the platform starts moving.
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