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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?
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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.
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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?.
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