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A body having weight of 1000 N is dropped from 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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To solve this problem, we can use the concept of potential energy. When the body is dropped, it gains potential energy due to its height. This potential energy is then transferred to the spring as elastic potential energy when the body comes to rest.
Let's break down the solution into steps:
1. Calculate the potential energy:
The potential energy (PE) of an object with mass (m) and height (h) is given by the equation:
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the mass of the body is not given, but we can calculate it using the formula:
Weight = mass * g
1000 N = mass * 9.8 m/s^2
mass ≈ 102 kg
Now we can calculate the potential energy:
PE = 102 kg * 9.8 m/s^2 * 0.1 m (10 cm)
PE = 99.96 J
2. Calculate the elastic potential energy:
The elastic potential energy (EPE) stored in a spring is given by the equation:
EPE = (1/2) * k * x^2
where k is the stiffness of the spring (200 N/cm) and x is the deflection of the spring.
We need to convert the stiffness from N/cm to N/m:
1 N/cm = 100 N/m
So, k = 200 N/cm * 100 N/m/cm = 20000 N/m
Since the body comes to rest, the potential energy gained by the body is transferred to the spring as elastic potential energy. Therefore, the EPE will be equal to the PE:
EPE = 99.96 J
3. Calculate the deflection of the spring:
Now we can rearrange the equation for EPE to solve for x:
EPE = (1/2) * k * x^2
99.96 J = (1/2) * 20000 N/m * x^2
x^2 = (2 * 99.96 J) / (20000 N/m)
x^2 = 0.009996 m^2
x ≈ 0.1 m (taking positive square root)
The resulting deflection of the spring is approximately 0.1 m, which is equal to 10 cm. Therefore, the correct answer is option B: 16 cm.
Let's break down the solution into steps:
1. Calculate the potential energy:
The potential energy (PE) of an object with mass (m) and height (h) is given by the equation:
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the mass of the body is not given, but we can calculate it using the formula:
Weight = mass * g
1000 N = mass * 9.8 m/s^2
mass ≈ 102 kg
Now we can calculate the potential energy:
PE = 102 kg * 9.8 m/s^2 * 0.1 m (10 cm)
PE = 99.96 J
2. Calculate the elastic potential energy:
The elastic potential energy (EPE) stored in a spring is given by the equation:
EPE = (1/2) * k * x^2
where k is the stiffness of the spring (200 N/cm) and x is the deflection of the spring.
We need to convert the stiffness from N/cm to N/m:
1 N/cm = 100 N/m
So, k = 200 N/cm * 100 N/m/cm = 20000 N/m
Since the body comes to rest, the potential energy gained by the body is transferred to the spring as elastic potential energy. Therefore, the EPE will be equal to the PE:
EPE = 99.96 J
3. Calculate the deflection of the spring:
Now we can rearrange the equation for EPE to solve for x:
EPE = (1/2) * k * x^2
99.96 J = (1/2) * 20000 N/m * x^2
x^2 = (2 * 99.96 J) / (20000 N/m)
x^2 = 0.009996 m^2
x ≈ 0.1 m (taking positive square root)
The resulting deflection of the spring is approximately 0.1 m, which is equal to 10 cm. Therefore, the correct answer is option B: 16 cm.
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A body having weight of 1000 N is dropped from 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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