NEET Exam > NEET Questions > A block of mass 15 kg is placed on a long tro...
Start Learning for Free
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?
Most Upvoted Answer
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 .
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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 detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup