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On stretching a spring by 0.1 m, a elastic potential energy of 5 J is stored in it. The force constant of the spring is
- a)100 N-m⁻¹
- b)250 N-m⁻¹
- c)500 N-m⁻¹
- d)1000 N-m⁻¹
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
On stretching a spring by 0.1 m, a elastic potential energy of 5 J is ...
X=0.l m
change in potential energy = 5j
spring constant = K
1/2 mV²=P.E = 1/2 *K *(0.1)²= 5
K = 1000 N/m
change in potential energy = 5j
spring constant = K
1/2 mV²=P.E = 1/2 *K *(0.1)²= 5
K = 1000 N/m
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On stretching a spring by 0.1 m, a elastic potential energy of 5 J is ...
The elastic potential energy stored in a spring can be calculated using the formula:
Elastic Potential Energy (PE) = (1/2) * k * x^2
Where:
k is the force constant of the spring
x is the displacement or stretch of the spring
Given that the elastic potential energy is 5 J and the stretch of the spring is 0.1 m, we can rearrange the formula to solve for k:
5 J = (1/2) * k * (0.1 m)^2
Simplifying the equation, we get:
5 J = (1/2) * k * 0.01 m^2
10 J = k * 0.01 m^2
Dividing both sides of the equation by 0.01 m^2, we find:
k = 10 J / 0.01 m^2
k = 1000 N/m
Therefore, the force constant of the spring is 1000 N/m.
Elastic Potential Energy (PE) = (1/2) * k * x^2
Where:
k is the force constant of the spring
x is the displacement or stretch of the spring
Given that the elastic potential energy is 5 J and the stretch of the spring is 0.1 m, we can rearrange the formula to solve for k:
5 J = (1/2) * k * (0.1 m)^2
Simplifying the equation, we get:
5 J = (1/2) * k * 0.01 m^2
10 J = k * 0.01 m^2
Dividing both sides of the equation by 0.01 m^2, we find:
k = 10 J / 0.01 m^2
k = 1000 N/m
Therefore, the force constant of the spring is 1000 N/m.
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On stretching a spring by 0.1 m, a elastic potential energy of 5 J is stored in it. The force constant of the spring isa)100 N-m¹b)250 N-m¹c)500 N-m¹d)1000 N-m¹Correct answer is option 'D'. Can you explain this answer?
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