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A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?
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A 10kg block is gently put on a moving plank with velocity 1 m/s on a ...
Work done on a block on a moving plank


Given information:


  • Mass of block (m) = 10 kg

  • Initial velocity of plank (u) = 1 m/s

  • Final velocity of plank (v) = 1 m/s

  • Coefficient of friction between plank and block (μ) = 0.4

  • Frictional force (f) = μmg



Explanation:

When the block is gently put on the moving plank, both the block and the plank will have the same velocity of 1 m/s. Due to the friction between the block and the plank, the block will start moving with the plank after some time. The external force that is maintaining the constant speed of the plank is equal to the frictional force acting on the block in the opposite direction.


The work done by the external force is equal to the product of the force and the displacement. As the plank is moving at a constant speed, the displacement is zero. Therefore, the work done by the external force is zero.


However, the work done by the frictional force between the block and the plank is not zero. The frictional force is given by f = μmg, where m is the mass of the block and g is the acceleration due to gravity. The work done by the frictional force is equal to the product of the force and the displacement, which is equal to the product of the force and the distance traveled by the block.


As the plank is moving at a constant speed, the distance traveled by the block is equal to the displacement of the plank. The displacement of the plank is given by (v-u)t, where t is the time for which the block is in contact with the plank. Therefore, the work done by the frictional force is given by:


Work done by frictional force = f(v-u)t = μmg(v-u)t = 0.4 x 10 x 9.8 x (1-1) x t = 0 Joules


Therefore, the work done by the external force is zero and the work done by the frictional force is also zero. Hence, the total work done on the block is zero.
Community Answer
A 10kg block is gently put on a moving plank with velocity 1 m/s on a ...
Is it -50N ? (taking g = 10m/s^2)
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A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?
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A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kg block is gently put on a moving plank with velocity 1 m/s on a frictionless horizontal ground surface and in due time the block starts with the plank at 1m/s the work done by the external force which maintains the constant speed of plank (coefficient of friction between plank and block is 0.4)?.
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