Summary:
This is lecture 5 for the topic Work, Energy and Power. You will find all the resource for the topic “Work, Energy and Power” by clicking this link.
1. Hooke’s Law and the Spring Force
An ideal spring is one that perfectly obeys Hooke’s Law. When a spring is stretched or compressed from its natural (relaxed) length, it exerts a restoring force that is directly proportional to the displacement.
Let the natural position of the free end of the spring be $x = 0$. If the free end is displaced to a position $x$, the spring force $F_s$ is given by:
$$F_s = -kx$$
- $k$ is the Spring Constant (or force constant), measured in $\text{N/m}$. It represents the stiffness of the spring. A higher $k$ means a stiffer spring.
- $x$ is the displacement from the natural length.
- The negative sign indicates that the spring force is a restoring force—it always acts in the direction opposite to the displacement, trying to bring the system back to equilibrium ($x=0$).
2. Work Done by a Spring Force
Since the spring force changes continuously with position (it is a variable force), we cannot use the simple $W = F \cdot d$ formula. We must integrate.
If a block attached to a spring is displaced from an initial position $x_i$ to a final position $x_f$, the work done by the spring is:
$$W_s = \int_{x_i}^{x_f} (-kx) dx = -k \left[ \frac{x^2}{2} \right]_{x_i}^{x_f}$$
$$W_s = \frac{1}{2}kx_i^2 – \frac{1}{2}kx_f^2$$
Important Implications:
- If the spring is stretched from its natural length ($x_i = 0$) to a position $x$, the work done by the spring is negative: $W_s = -\frac{1}{2}kx^2$.
- The work done by the spring only depends on the initial and final positions, not the path taken. Therefore, the spring force is a conservative force.
3. Elastic Potential Energy of a Spring ($U_s$)
Because the spring force is conservative, we can define a potential energy function for it. The change in potential energy is the negative of the work done by the conservative force: $\Delta U = -W_s$.
Taking the natural length of the spring ($x=0$) as the reference point where potential energy is zero ($U_i = 0$), the potential energy $U$ at any displacement $x$ is:
$$U_s(x) = \frac{1}{2}kx^2$$
Key Characteristics of Spring PE:
- It is always positive or zero ($U_s \ge 0$), regardless of whether the spring is stretched ($x > 0$) or compressed ($x < 0$).
- The $U-x$ graph is an upward-opening parabola with its vertex at the origin. This shape mathematically guarantees that $x=0$ is a point of stable equilibrium.
4. Combinations of Springs
In JEE problems, springs are often combined. You can replace a combination of springs with a single “equivalent spring” having a constant $k_{eq}$.
A. Series Combination
When springs are connected end-to-end, the tension (force $F$) is the same in all springs, but their total extension is the sum of their individual extensions ($x = x_1 + x_2 + \dots$).
$$\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \dots + \frac{1}{k_n}$$
Tip: If an ideal spring of constant $k$ and length $L$ is cut into two equal halves, each half will have a spring constant of $2k$.
B. Parallel Combination
When springs are connected side-by-side between the same two blocks/walls, they share the same extension ($x$), but the total force is the sum of the individual forces ($F = F_1 + F_2 + \dots$).
$$k_{eq} = k_1 + k_2 + \dots + k_n$$
JEE-Mains/NEET Practice Questions
Question 1: Work Done on a Spring
A spring with a force constant $k = 50 \text{ N/m}$ is initially stretched by $0.1 \text{ m}$ from its natural length. What is the external work required to stretch it further to a total displacement of $0.3 \text{ m}$?
Solution:
The work done by an external agent is equal to the change in the potential energy of the spring (assuming it is stretched slowly, $\Delta K = 0$).
$W_{ext} = \Delta U = U_f – U_i$
$W_{ext} = \frac{1}{2}k x_f^2 – \frac{1}{2}k x_i^2 = \frac{1}{2}k(x_f^2 – x_i^2)$
$W_{ext} = \frac{1}{2}(50)(0.3^2 – 0.1^2) = 25(0.09 – 0.01) = 25(0.08)$
$W_{ext} = 2.0 \text{ J}$
Question 2: Cutting a Spring
An ideal spring of natural length $L$ and spring constant $k$ is cut into two pieces. The lengths of the pieces are in the ratio $1 : 2$. What is the spring constant of the longer piece?
Solution:
The spring constant $k$ of a uniform spring is inversely proportional to its natural length $l$ ($k \propto 1/l$).
Therefore, $k \cdot l = \text{constant}$.
The two pieces have lengths $l_1 = \frac{L}{3}$ and $l_2 = \frac{2L}{3}$.
Let $k_2$ be the spring constant of the longer piece.
$k \cdot L = k_2 \cdot l_2 \implies k \cdot L = k_2 \cdot \left(\frac{2L}{3}\right)$
$k_2 = \frac{3k}{2}$
Question 3: Energy Conservation with a Spring
A block of mass $m = 2 \text{ kg}$ is moving on a frictionless horizontal surface with a velocity of $4 \text{ m/s}$. It collides with a horizontal, massless spring of spring constant $k = 800 \text{ N/m}$. What is the maximum compression of the spring?
Solution:
Apply conservation of mechanical energy. Initially, the energy is entirely kinetic. At maximum compression ($x_m$), the block stops momentarily ($v = 0$), so the energy is entirely elastic potential energy.
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kx_m^2$
$x_m = \sqrt{\frac{m}{k}} v = \sqrt{\frac{2}{800}} \times 4 = \sqrt{\frac{1}{400}} \times 4$
$x_m = \frac{1}{20} \times 4 = 0.2 \text{ m}$ (or $20 \text{ cm}$)
Question 4: Vertical Spring System
A mass $m$ is gently lowered onto an unstretched vertical spring of constant $k$ until it reaches its new equilibrium position. How much work is done by gravity during this process, and what is the change in the spring’s potential energy?
Solution:
- At the new equilibrium position, the upward spring force equals the downward gravitational force: $kx_0 = mg \implies x_0 = \frac{mg}{k}$.
- The mass moved down by $x_0$. Work done by gravity: $W_g = mgx_0 = mg\left(\frac{mg}{k}\right) = \frac{(mg)^2}{k}$.
- The change in the spring’s potential energy: $\Delta U_s = \frac{1}{2}kx_0^2 = \frac{1}{2}k\left(\frac{mg}{k}\right)^2 = \frac{(mg)^2}{2k}$.
(Note: The remaining energy, $\frac{(mg)^2}{2k}$, was removed by the external agent that “gently lowered” the mass. If it were dropped suddenly, it would oscillate!)