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A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force?
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A block of mass 15 kg is placed on a long trolley , The coefficient of...
Problem: Find the friction force acting on a block of mass 15 kg placed on a long trolley accelerating from rest with 0.5 m/s for 20 s, given that the coefficient of friction between the block and trolley is 0.18.
Solution:
Therefore, the friction force acting on the block of mass 15 kg placed on a long trolley accelerating from rest with 0.5 m/s for 20 s, given that the coefficient of friction between the block and trolley is 0.18, is 26.47 N.
Solution:
- Acceleration of the trolley: The given acceleration of the trolley is 0.5 m/s for 20 s. Using the formula of acceleration, we get:
- a = (v - u) / t
- a = (0.5 - 0) / 20
- a = 0.025 m/s²
- Force acting on the trolley: The force acting on the trolley can be calculated using Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. Therefore, we get:
- F = m * a
- F = 15 * 0.025
- F = 0.375 N
- Friction force: The friction force can be calculated using the formula of friction force, which is equal to the coefficient of friction multiplied by the normal force. Therefore, we get:
- f = μ * N
- f = 0.18 * 15 * 9.81
- f = 26.47 N
Therefore, the friction force acting on the block of mass 15 kg placed on a long trolley accelerating from rest with 0.5 m/s for 20 s, given that the coefficient of friction between the block and trolley is 0.18, is 26.47 N.
Community Answer
A block of mass 15 kg is placed on a long trolley , The coefficient of...
Given,mass of block ( Mb) = 15 kgCoefficient of friction between the block and the trolley (u) = 0.18acceleration of the trolley = 0.5 m/s^2Block is placed on the trolley , so force applied on the block by the trolleyF = Mb*a= 15 * 0.5 N= 7.5 N this is reaction force applied by the trolley on the block , so its direction is opposite of motion of the trolley .hence limiting friction oppose the motion of the block .we know ,body will be rest when limiting friction is greater then applied force on the block by the trolley .now,Limiting Friction (f) = uN= 0.18 * Mb*g= 0.18 * 15* 9.8= 26.46 Nhere, f> F , so static friction adjusr itself .hence, block will remain at rest on the trolley .after 20 sec trolley moves with uniform velocity . so, acceleration and force applied by trolley on the block will be zero. it means there is no frictional force act on the block .e.g block appear at rest relative to trolley to a STATIONARY observer on the ground .
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A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force?
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A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force?.
A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 15 kg is placed on a long trolley , The coefficient of friction between block and trolley is 0.18 . The trolley accelerates from rest with 0.5 m/s for 20 s , then what is the friction force?.
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