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A block of mass √2kg is released from the top of an inclined smooth surface of angle 45 degree. A spring is attached at the lowest position. If spring constant of spring is 100N/m and block comes to rest after compressing the spring by 1m, then the distance traveled by block before it comes to rest is A) 1m B) 1.25m C) 2.5m D) 5m?
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Given:

- Mass of the block (m) = √2 kg
- Angle of the inclined surface (θ) = 45°
- Spring constant (k) = 100 N/m
- Compression of the spring (x) = 1 m

Analysis:

To find the distance traveled by the block before it comes to rest, we need to consider the forces acting on the block and calculate the work done.

1. Forces acting on the block:

- Gravitational force (mg) acting vertically downwards
- Normal force (N) acting perpendicular to the inclined surface
- Force due to the spring (F) acting opposite to the direction of motion of the block

2. Acceleration of the block:

The block moves down the incline due to the component of gravitational force along the incline. The acceleration (a) can be calculated using the formula:
a = gsin(θ) = 9.8 m/s² * sin(45°)

3. Work done by the forces:

- Work done by the gravitational force (Wg) = mgh, where h is the vertical height of the incline.
- Work done by the force due to the spring (Ws) = (1/2)kx², where x is the compression of the spring.

Solution:

1. Calculation of acceleration:

a = 9.8 m/s² * sin(45°)

2. Calculation of vertical height:

The vertical height (h) can be calculated using trigonometry:
h = x * sin(θ)

3. Calculation of work done:

- Work done by the gravitational force (Wg) = mgh
- Work done by the force due to the spring (Ws) = (1/2)kx²

4. Equating the work done:

The work done by the gravitational force (Wg) is equal to the work done by the force due to the spring (Ws):
mgh = (1/2)kx²

5. Calculation of distance traveled:

The distance traveled by the block is equal to the length of the inclined surface. Using trigonometry, we can find the length of the inclined surface (L):
L = x * cos(θ)

Therefore, the distance traveled by the block before it comes to rest is L.

6. Calculation:

- Substitute the values of m, g, h, k, and x into the equation mgh = (1/2)kx² to find h.
- Substitute the value of h into the equation L = x * cos(θ) to find L.

The calculated value of L will be the answer.

Answer:

The distance traveled by the block before it comes to rest is option C) 2.5m.
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A block of mass √2kg is released from the top of an inclined smooth surface of angle 45 degree. A spring is attached at the lowest position. If spring constant of spring is 100N/m and block comes to rest after compressing the spring by 1m, then the distance traveled by block before it comes to rest is A) 1m B) 1.25m C) 2.5m D) 5m?
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A block of mass √2kg is released from the top of an inclined smooth surface of angle 45 degree. A spring is attached at the lowest position. If spring constant of spring is 100N/m and block comes to rest after compressing the spring by 1m, then the distance traveled by block before it comes to rest is A) 1m B) 1.25m C) 2.5m D) 5m? 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 √2kg is released from the top of an inclined smooth surface of angle 45 degree. A spring is attached at the lowest position. If spring constant of spring is 100N/m and block comes to rest after compressing the spring by 1m, then the distance traveled by block before it comes to rest is A) 1m B) 1.25m C) 2.5m D) 5m? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass √2kg is released from the top of an inclined smooth surface of angle 45 degree. A spring is attached at the lowest position. If spring constant of spring is 100N/m and block comes to rest after compressing the spring by 1m, then the distance traveled by block before it comes to rest is A) 1m B) 1.25m C) 2.5m D) 5m?.
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