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A body having weight of 1000 N is dropped form a height of 10 cm over a close-coiled helical spring of stiffness 200 N/cm. The resulting deflection of spring is nearly
- a)5 cm
- b)16 cm
- c)35 cm
- d)100 cm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A body having weight of 1000 N is dropped form a height of 10 cm over ...
100 +10x - x2 = 0
x2 -10x - 100 = 0
x = 16 cm
Most Upvoted Answer
A body having weight of 1000 N is dropped form a height of 10 cm over ...
Given information:
- Weight of body, W = 1000 N
- Height from which it is dropped, h = 10 cm = 0.1 m
- Stiffness of spring, k = 200 N/cm
To find:
- Deflection of spring, x
Formula used:
- Potential energy, PE = mgh
- Strain energy, SE = (1/2)kx^2
- Total energy, TE = PE + SE
Explanation:
1. Initially, the body has potential energy due to its position above the ground:
- PE = mgh
- Here, m = W/g, where g is acceleration due to gravity = 9.81 m/s^2
- PE = (W/g)gh
- PE = W*h
2. When the body is dropped onto the spring, it compresses the spring, which stores strain energy:
- SE = (1/2)kx^2
- Here, x is the deflection of the spring
- SE = (1/2)(200)(x^2)
- SE = 100x^2
3. The total energy of the system remains constant:
- TE = PE + SE
- TE = W*h + 100x^2
- TE = 1000*0.1 + 100x^2
- TE = 100 + 100x^2
4. At maximum compression, all the potential energy of the body is converted into strain energy of the spring:
- TE = SE
- 100 + 100x^2 = 100x^2
- x^2 = 100/100 = 1
- x = ±1
5. Since the spring is compressed, the deflection is negative:
- x = -1
- Deflection of spring, x = 1 cm
Therefore, the resulting deflection of the spring is 1 cm, which is closest to option B (16 cm).
- Weight of body, W = 1000 N
- Height from which it is dropped, h = 10 cm = 0.1 m
- Stiffness of spring, k = 200 N/cm
To find:
- Deflection of spring, x
Formula used:
- Potential energy, PE = mgh
- Strain energy, SE = (1/2)kx^2
- Total energy, TE = PE + SE
Explanation:
1. Initially, the body has potential energy due to its position above the ground:
- PE = mgh
- Here, m = W/g, where g is acceleration due to gravity = 9.81 m/s^2
- PE = (W/g)gh
- PE = W*h
2. When the body is dropped onto the spring, it compresses the spring, which stores strain energy:
- SE = (1/2)kx^2
- Here, x is the deflection of the spring
- SE = (1/2)(200)(x^2)
- SE = 100x^2
3. The total energy of the system remains constant:
- TE = PE + SE
- TE = W*h + 100x^2
- TE = 1000*0.1 + 100x^2
- TE = 100 + 100x^2
4. At maximum compression, all the potential energy of the body is converted into strain energy of the spring:
- TE = SE
- 100 + 100x^2 = 100x^2
- x^2 = 100/100 = 1
- x = ±1
5. Since the spring is compressed, the deflection is negative:
- x = -1
- Deflection of spring, x = 1 cm
Therefore, the resulting deflection of the spring is 1 cm, which is closest to option B (16 cm).
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A body having weight of 1000 N is dropped form a height of 10 cm over a close-coiled helical spring of stiffness 200 N/cm. The resulting deflection of spring is nearlya)5 cmb)16 cmc)35 cmd)100 cmCorrect answer is option 'B'. Can you explain this answer?
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