
Two springs have the spring constants ${k_A}$ and ${k_B}$ and ${k_A} > {k_B}$. The work required to stretch them by the same extension will be:
a. More in spring A
b. More in spring B
c. Equal in both
d. Nothing can be said
Answer
560.4k+ views
Hint:
Work done depends on both spring constant and displacement of the spring directly. In fact, work done is directly proportional to both. So, Higher the spring constant lower the displacement of spring.
Formula used:
1. Work Done, ${W_{\text{spring}}} = \dfrac{1}{2}{kx_{\text{final}}}^2$ …… (1)
Where,
${W_{\text{spring}}}$ is work done by spring
k is spring constant
${x_{\text{final}}}$is net displacement from mean position.
2. Spring force $= F = - kx$ …… (2)
Where,
F is force acting on body
k is spring constant
$x$net displacement from the mean position of the body.
Complete step by step answer:
Given,
1. Spring constants,${k_A}$ and ${k_B}$ for two springs.
2. ${k_A} > {k_B}$ …… (3)
3. $\Delta {x_A} = \Delta {x_B}$ …… (4)
Step 1:
Using equation (4), let’s say $\Delta {x_A} = \Delta {x_B} = x$ (same for both)
Step 2:
Using equation (1), (5) and (6), we can find work done for displacement x in for both setups.
Work done in expanding A spring setup to x displacement:
${W_A} = \dfrac{1}{2}{k_A}{x^2}$ …… (7)
Similarly,
Work done in expanding B spring setup to x displacement:
${W_B} = \dfrac{1}{2}{k_B}{x^2}$ …… (8)
Step 3:
From equation (3) we know, ${k_A} > {k_B}$
So, comparing equation (7) and (8) we can say
${W_A} > {W_B}$
Hence, the correct answer is option (B).
Note: Another method to solve
Step 1:
Since x (displacement of spring) is the same for both springs. So, Work done would be just dependent rather proportional to spring constant. $W \propto k$ where, k is spring constant and W is work.
Step 2:
So, Higher the spring constant of spring more would be work done for moving by same displacement.
From equation (3) we know
${k_A} > {k_B} \Rightarrow {W_A} > {W_B}$.
Work done depends on both spring constant and displacement of the spring directly. In fact, work done is directly proportional to both. So, Higher the spring constant lower the displacement of spring.
Formula used:
1. Work Done, ${W_{\text{spring}}} = \dfrac{1}{2}{kx_{\text{final}}}^2$ …… (1)
Where,
${W_{\text{spring}}}$ is work done by spring
k is spring constant
${x_{\text{final}}}$is net displacement from mean position.
2. Spring force $= F = - kx$ …… (2)
Where,
F is force acting on body
k is spring constant
$x$net displacement from the mean position of the body.
Complete step by step answer:
Given,
1. Spring constants,${k_A}$ and ${k_B}$ for two springs.
2. ${k_A} > {k_B}$ …… (3)
3. $\Delta {x_A} = \Delta {x_B}$ …… (4)
Step 1:
Using equation (4), let’s say $\Delta {x_A} = \Delta {x_B} = x$ (same for both)
Step 2:
Using equation (1), (5) and (6), we can find work done for displacement x in for both setups.
Work done in expanding A spring setup to x displacement:
${W_A} = \dfrac{1}{2}{k_A}{x^2}$ …… (7)
Similarly,
Work done in expanding B spring setup to x displacement:
${W_B} = \dfrac{1}{2}{k_B}{x^2}$ …… (8)
Step 3:
From equation (3) we know, ${k_A} > {k_B}$
So, comparing equation (7) and (8) we can say
${W_A} > {W_B}$
Hence, the correct answer is option (B).
Note: Another method to solve
Step 1:
Since x (displacement of spring) is the same for both springs. So, Work done would be just dependent rather proportional to spring constant. $W \propto k$ where, k is spring constant and W is work.
Step 2:
So, Higher the spring constant of spring more would be work done for moving by same displacement.
From equation (3) we know
${k_A} > {k_B} \Rightarrow {W_A} > {W_B}$.
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