Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with a same force. Their potential energies will be in the ratio of
Potential energy is the stored energy of an object due to its position or configuration. Common types include gravitational potential energy dependent on height and mass and elastic potential energy associated with the deformation of elastic materials. It plays a crucial role in energy conservation and mechanical systems.
Chapter 5 of Class 11 Physics discusses work energy and power. It defines work done by a force energy conservation and transformations as well as types of energy including kinetic and potential. The chapter also introduces power as the rate of doing work highlighting its significance in various applications.
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To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension of the spring
Step 1: Relationship between force and extension
According to Hooke’s Law, the force applied to a spring is related to the spring constant and the extension by:
F = kx
Rearranging gives us:
x = F/k
Step 2: Calculate potential energy for both springs
For spring 1 (k₁ = 1500 N/m):
U₁ = (1/2) k₁ x₁²
Substituting for x₁:
U₁ = (1/2) k₁ (F/k₁)²
U₁ = (1/2) k₁ (F²/k₁²)
U₁ = (1/2) (F²/k₁)
For spring 2 (k₂ = 3000 N/m):
U₂ = (1/2) k₂ x₂²
Substituting for x₂:
U₂ = (1/2) k₂ (F/k₂)²
U₂ = (1/2) k₂ (F²/k₂²)
U₂ = (1/2) (F²/k₂)
Step 3: Find the ratio of potential energies
Now we can find the ratio of U₁ to U₂:
U₁ / U₂ = (F² / (2k₁)) / (F² / (2k₂))
U₁ / U₂ = k₂ / k₁
U₁ / U₂ = 3000 / 1500
U₁ / U₂ = 2
Final Answer:
The potential energies of the two springs will be in the ratio of 2 : 1.
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