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A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shownin the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)Q.4.The magnitude of the normal reaction that acts on the block at the point Q isa)7.5Nb)8.6Nc)11.5Nd)22.5NCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared
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the JEE exam syllabus. Information about A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shownin the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)Q.4.The magnitude of the normal reaction that acts on the block at the point Q isa)7.5Nb)8.6Nc)11.5Nd)22.5NCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam.
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Here you can find the meaning of A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shownin the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)Q.4.The magnitude of the normal reaction that acts on the block at the point Q isa)7.5Nb)8.6Nc)11.5Nd)22.5NCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shownin the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)Q.4.The magnitude of the normal reaction that acts on the block at the point Q isa)7.5Nb)8.6Nc)11.5Nd)22.5NCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shownin the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)Q.4.The magnitude of the normal reaction that acts on the block at the point Q isa)7.5Nb)8.6Nc)11.5Nd)22.5NCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shownin the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)Q.4.The magnitude of the normal reaction that acts on the block at the point Q isa)7.5Nb)8.6Nc)11.5Nd)22.5NCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shownin the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)Q.4.The magnitude of the normal reaction that acts on the block at the point Q isa)7.5Nb)8.6Nc)11.5Nd)22.5NCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.