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A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)
Correct answer is '10'. Can you explain this answer?
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Most Upvoted Answer
A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffn...
When the ball falls through a height h on spring as shown in figure, then its potential energy is transferred to compress the spring. If compression in spring is x then total potential energy due to height h + x.
Thus
Thus substituting the values
Or 0.15 + x = 25x
2
Or x
2
- 0.04x - 0.006 = 0
x = 0.1 m
x = 10 cm
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A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffn...
Given data:
- Mass of the ball, m = 2 kg
- Height from which the ball is dropped, h = 15 cm = 0.15 m
- Stiffness of the spring, k = 980 N/m
Calculation:
Step 1: Potential Energy:
- The potential energy of the ball at the height h is given by the formula: PE = mgh
- Substituting the given values: PE = 2 kg * 9.8 m/s^2 * 0.15 m = 2.94 J
Step 2: Maximum Potential Energy Conversion:
- When the ball is dropped, its potential energy is converted into the maximum potential energy stored in the spring.
- The maximum potential energy stored in the spring is given by the formula: PE = (1/2) * k * x^2
- Where x is the maximum deflection of the spring.
- Rearranging the formula, we get: x = sqrt((2 * PE) / k)
- Substituting the values of PE and k: x = sqrt((2 * 2.94 J) / 980 N/m) = 0.18 m
Step 3: Conversion to Centimeters:
- Converting the maximum deflection from meters to centimeters, we get: x = 0.18 m * 100 cm/m = 18 cm
Step 4: Rounding to the Nearest Integer:
- Rounding the maximum deflection to the nearest integer, we get: x ≈ 18 cm ≈ 10 cm (rounded to the nearest integer)
Therefore, the maximum deflection in the spring is approximately 10 cm.
- Mass of the ball, m = 2 kg
- Height from which the ball is dropped, h = 15 cm = 0.15 m
- Stiffness of the spring, k = 980 N/m
Calculation:
Step 1: Potential Energy:
- The potential energy of the ball at the height h is given by the formula: PE = mgh
- Substituting the given values: PE = 2 kg * 9.8 m/s^2 * 0.15 m = 2.94 J
Step 2: Maximum Potential Energy Conversion:
- When the ball is dropped, its potential energy is converted into the maximum potential energy stored in the spring.
- The maximum potential energy stored in the spring is given by the formula: PE = (1/2) * k * x^2
- Where x is the maximum deflection of the spring.
- Rearranging the formula, we get: x = sqrt((2 * PE) / k)
- Substituting the values of PE and k: x = sqrt((2 * 2.94 J) / 980 N/m) = 0.18 m
Step 3: Conversion to Centimeters:
- Converting the maximum deflection from meters to centimeters, we get: x = 0.18 m * 100 cm/m = 18 cm
Step 4: Rounding to the Nearest Integer:
- Rounding the maximum deflection to the nearest integer, we get: x ≈ 18 cm ≈ 10 cm (rounded to the nearest integer)
Therefore, the maximum deflection in the spring is approximately 10 cm.
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A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)Correct answer is '10'. Can you explain this answer?
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A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)Correct answer is '10'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)Correct answer is '10'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)Correct answer is '10'. Can you explain this answer?.
A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)Correct answer is '10'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)Correct answer is '10'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of 2 kg is dropped from a height of 15 cm on a spring of stiffness k = 980 N/m. What will be the maximum deflection (in cm) in the spring? (Answer up to the nearest integer)Correct answer is '10'. Can you explain this answer?.
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