Question

Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identical weights are hooked to their ends, and the weights are allowed to stretch the springs. The ratio of the energy stored by spring #1 to that stored by spring #2 is

1:1

1:4

2:1

1:2

Answer 1

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Step-by-step explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by
`(1)/(2)kx^(2)` where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is,
`(1)/(2)k(x1)^(2)`

⇒
`(1)/(2)k((w)/(k))^(2)`

⇒
`(w^(2))/(2k)`

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is,
`(1)/(2)2k(x2)^(2)`

⇒
`(1)/(2)2k((w)/(2k))^(2)`

⇒
`(w^(2))/(4k)`

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is
`((w^(2))/(2k))/((w^(2))/(4k))=`2:1