JEE Exam > JEE Questions > A block of mass 0.18 kg is attached to a spri...
Start Learning for Free
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:
Correct answer is '4'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/...
Let ν be the speed of the block just after impulse. At B, the block comes to rest.
Therefore: Loss in K.E. of the block = Gain in P.E. of the spring + Work done against friction
Most Upvoted Answer
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/...
Initial Conditions:
- Mass of the block, m = 0.18 kg
- Force constant of the spring, k = 2 N/m
- Coefficient of friction between the block and the floor, μ = 0.1
- Distance the block slides, x = 0.06 m
- Initial velocity of the block, V = N/10
To find the value of N, we need to analyze the motion of the block and determine the work done by the impulse.
1. Finding the Spring Constant:
The force exerted by the spring can be given by Hooke's Law as F = -kx, where x is the displacement from the equilibrium position.
Since the block is at rest initially, the spring is unstretched, and the net force is zero. Therefore, the force due to the impulse must be balanced by the force exerted by the spring.
Impulse = Force × Time
mV = kx × t
t = mx / k
2. Determining the Friction Force:
The block slides a distance of x = 0.06 m and comes to rest due to the friction force.
The work done by the friction force can be calculated using the formula:
Work = Force × Distance
μmgx = (μmg) × x
3. Equating Work Done:
Since the block comes to rest, the work done by the impulse is equal to the work done by the friction force.
mV^2 / 2 = μmgx
V^2 = 2μgx
(N/10)^2 = 2(0.1)(0.18)(0.06)
N^2 = 0.00216
N ≈ √0.00216 ≈ 0.0465
4. Final Answer:
N ≈ 0.0465
N ≈ 4.
- Mass of the block, m = 0.18 kg
- Force constant of the spring, k = 2 N/m
- Coefficient of friction between the block and the floor, μ = 0.1
- Distance the block slides, x = 0.06 m
- Initial velocity of the block, V = N/10
To find the value of N, we need to analyze the motion of the block and determine the work done by the impulse.
1. Finding the Spring Constant:
The force exerted by the spring can be given by Hooke's Law as F = -kx, where x is the displacement from the equilibrium position.
Since the block is at rest initially, the spring is unstretched, and the net force is zero. Therefore, the force due to the impulse must be balanced by the force exerted by the spring.
Impulse = Force × Time
mV = kx × t
t = mx / k
2. Determining the Friction Force:
The block slides a distance of x = 0.06 m and comes to rest due to the friction force.
The work done by the friction force can be calculated using the formula:
Work = Force × Distance
μmgx = (μmg) × x
3. Equating Work Done:
Since the block comes to rest, the work done by the impulse is equal to the work done by the friction force.
mV^2 / 2 = μmgx
V^2 = 2μgx
(N/10)^2 = 2(0.1)(0.18)(0.06)
N^2 = 0.00216
N ≈ √0.00216 ≈ 0.0465
4. Final Answer:
N ≈ 0.0465
N ≈ 4.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer?
Question Description
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer?.
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer?.
Solutions for A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer?, a detailed solution for A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? has been provided alongside types of A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block m/s is V = N/10. Then N is:Correct answer is '4'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE 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
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup