Question

You have two springs that are identical except that spring 1 is stiffer than spring 2, (k1 > k2). On which spring is more work done?(a) If they are stretched using the same force(b) If they are stretched the same distance

Answer 1

If stretched using the same force, more work is done on the less stiff spring (spring 2). If stretched the same distance, more work is done on the stiffer spring (spring 1) due to its higher spring constant.

Step-by-step explanation:

When evaluating the work done on two springs with different stiffness constants (k₁ and k₂ where k₁ > k₂), two scenarios are considered:

(a) Stretched using the same force

If both springs are stretched using the same force, the stiffer spring (spring 1) will have less extension than the more compliant spring (spring 2). Since work is the product of force and distance (W = F × x), the spring that stretches more (spring 2) will have more work done on it, assuming the force is constant and applied over a greater distance.

(b) Stretched the same distance

When both springs are stretched the same distance, the work done is given by the formula W = 1/2 × k × x². Given that k₁ > k₂, stretching spring 1 will require more work since the spring constant (k) is greater, and the distance (x) is the same for both.

Therefore, spring 1 will have more work done on it when stretched the same distance as spring 2.

Answer 2

Step-by-step explanation:

Given

Two springs with spring constant
`k_1` and
`k_2`

(a)If they are stretched using the same force

Force
`F=k_1x_1=k_2x_2`

where
`x_1` and
`x_2` are the extension in the spring

so
`(k_1)/(k_2)=(x_2)/(x_1)`

therefore
`x_2 > x_1`

Work done is given by

`W=F\cdot x`

`W_1=F\cdot x_1`

`W_2=F\cdot x_2`

thus
`W_2 > W_1`

(b)If they are stretched the same distance

`(F_1)/(F_2)=(k_1)/(k_2)`

Thus
`F_1 > F_2`

For same extension, work done by Force 1 is more

`W_1 > W_2`