This question has Statement I and Statement II. Of the four choices given after the statements, choo

Q: Why can't I just use $W = \frac{1}{2}kx^2$ for the first part?

You can, but it makes the comparison harder! If you use $W = \frac{1}{2}kx^2$, you must remember that $x$ is not constant; it depends on $k$ (since $x = F/k$). Substituting this back gives $W = \frac{F^2}{2k}$, which directly shows the relationship when $F$ is constant.

Q: Does a higher spring constant mean the spring does more work?

It depends entirely on the constraints! If you pull them by the same distance, the stiffer spring (higher $k$) requires more work. If you pull them with the same force, the softer spring (lower $k$) stretches much further, resulting in more work done on it.

Q: What does $k_1 < k_2$ physically mean?

It means spring 1 is "softer" or more compliant than spring 2. It takes less force to stretch spring 1 by the same amount compared to spring 2.

Q: Is the work done on the spring always equal to the potential energy stored?

Yes, as long as the spring is stretched slowly (quasi-statically) and there are no non-conservative forces like internal friction, the work done by the external force is entirely stored as elastic potential energy.

Q: Why did Statement I turn out to be false?

Statement I assumed that $W_1$ would still be greater than $W_2$ even when the displacement was the same. However, we proved $k_1 < k_2$. For a constant displacement, work is directly proportional to $k$, meaning $W_1$ must be less than $W_2$.

Question

This question has Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
If two springs $S_1$ and $S_2$ of force constants $k_1$ and $k_2$ respectively are stretched by the same force, it is found that more work is done on spring $S_1$ than on spring $S_2$.
**Statement I** If stretched by the same amount, work done on $S_1$ will be more than that on $S_2$.
**Statement II** $k_1 < k_2$

Options:

(A) Statement I is false, Statement II is true

(B) Statement I is true, Statement II is false

(C) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I

(D) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I

Sigma Percentile · Accepted Answer

### The Tale of Two Springs: Choosing the Right Work Formula Imagine you are in a physics lab holding two different springs, $S_1$ and $S_2$. One might be a thick suspension spring from a car, and the other might be a flimsy spring from a ballpoint pen. The problem presents us with a fascinating scenario: when you pull both springs with the exact same force $F$, you end up doing more work on spring $S_1$. How is this possible, and what does it tell us about the stiffness of these springs? ### Analyzing the Setup: The Constant Force Constraint To understand the first premise, we need to look at the formula for the work done on a spring. The most famous form is $W = \frac{1}{2}kx^2$. However, using this formula directly is tricky here because if the force is constant, the displacement $x$ will be different for each spring. Instead, let's substitute $x = \frac{F}{k}$ into the work equation: $$W = \frac{1}{2}k\left(\frac{F}{k} ight)^2 = \frac{F^2}{2k}$$ This is our **Master Equation** for constant force scenarios. It beautifully isolates the variables. Since the force $F$ is the same for both springs, the work done is inversely proportional to the spring constant $k$ ($W \propto \frac{1}{k}$). ### Evaluating Statement II The problem states that $W_1 > W_2$. Using our inverse relationship, if more work is done on $S_1$, it must have a smaller spring constant. $$\frac{F^2}{2k_1} > \frac{F^2}{2k_2} \implies k_1 < k_2$$ Physically, this makes perfect sense! Spring $S_1$ is "softer." When you pull it with force $F$, it stretches much further than the stiffer spring $S_2$. Since Work = Force $ imes$ Displacement, the larger displacement of $S_1$ results in more work being done. Therefore, **Statement II is absolutely true**. ### Evaluating Statement I: The Constant Displacement Constraint Now, the rules of the game change. Statement I asks us to imagine stretching both springs by the *same amount* (constant displacement $x$). Which formula should we use now? Since $x$ is constant, we return to the classic formula: $$W = \frac{1}{2}kx^2$$ In this scenario, the work done is directly proportional to the spring constant ($W \propto k$). We already proved that $k_1 < k_2$. Therefore, if we stretch them by the same amount, it will require less work to stretch the softer spring $S_1$. $$W_1 < W_2$$ Statement I claims that the work done on $S_1$ will be more, which contradicts our derivation. Thus, **Statement I is false**. ### Final Conclusion By carefully selecting the correct mathematical representation of work based on the physical constraints (constant force vs. constant displacement), we easily decoded the problem. Statement I is false, and Statement II is true, leading us confidently to option (a).

Select Answer:

Visualized Solution

$W_{1} > W_{2} for same F$

$W = \frac{F ^{2}}{2 k}$

$k_{1} < k_{2}$

$x_{1} = x_{2} = x$

$W = \frac{1}{2} k x^{2}$

$W_{1} < W_{2}$

\text{Conclusion}

The Sigma Insight: Work Done by Forces

The Tale of Two Springs

Analyzing the Setup

Evaluating Statement II

Evaluating Statement I

Final Conclusion

Similar Questions

A spring of spring constant $5 \times 1 0^{3}$ N/m is stretched initially by $5$ cm from the unstretched position. Then, the work required to stretch it further by another $5$ cm is

A spring of force constant $800 N/m$ has an extension of $5 cm$ . The work done in extending it from $5 cm$ to $15 cm$ is

When a rubber band is stretched by a distance $x$ , it exerts a restoring force of magnitude $F = a x + b x^{2}$ , where $a$ and $b$ are constants. The work done in stretching the unstretched rubber band by $L$ is

If $W_{1}$ , $W_{2}$ and $W_{3}$ represent the work done in moving a particle from $A$ to $B$ along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass $m$ . Find the correct relation between $W_{1}$ , $W_{2}$ and $W_{3}$

A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10000 N/m. The spring compresses by

Consider a force $F = - x \hat{i} + y \hat{j}$ . The work done by this force in moving a particle from point $A (1, 0)$ to $B (0, 1)$ along the line segment is (all quantities are in SI units)

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W1​>W2​ for same F

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=2kF2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k1​<k2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​=x2​=x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=21​kx2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W1​<W2​