Physics · Work, Energy And Power · NEET
It is 1/2 k x^2. The 1/2 appears because the spring force is NOT constant, it grows from 0 (at natural length) to k x (at stretch x). The work done to stretch it is the AREA of a force-displacement triangle: 1/2 * base * height = 1/2 * x * (k x) = 1/2 k x^2. If you wrongly used force = k x as constant, you would get k x^2, which is double the correct value. The 1/2 is the average factor for a linearly increasing force.
Because the restoring force itself increases in proportion to x (F = k x). Stretching further needs a bigger and bigger force each step. Adding up all that work gives an x-squared result. Practically this means: stretch 2 cm stores U; stretch 8 cm (4 times more) stores 4^2 = 16 times U. This exact ratio idea is tested directly in NEET 2023.
No. Whether you stretch (x positive) or compress (x negative), x^2 is always positive, so U = 1/2 k x^2 is always positive. NCERT states the spring force does work -k x^2/2 in both cases, and the stored elastic PE is +1/2 k x^2. A compressed spring stores just as much energy as a stretched spring of the same magnitude x.
x is always the displacement from the natural (unstretched, equilibrium) length of the spring, NOT from the floor or any other point. NCERT sets V(x) = 0 at x = 0, the equilibrium position. If a problem gives you total length, first subtract the natural length to get the true extension x before squaring it.
Force F = k x has units of newtons (N) and tells you how hard the spring pulls back at a given stretch. Energy U = 1/2 k x^2 has units of joules (J) and tells you how much work is stored. Force is the slope of the line on a force vs x graph; energy is the area under that line. NEET mixes these two, so read the question carefully to see whether it asks for force or energy.
The potential energy of a spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it will be
Try the real previous-year questions from this chapter — each with the answer and a full solution.
U = 1/2 k x^2, where k is the spring constant (N/m) and x is the displacement from the natural length (m). The unit of U is the joule (J).
The spring force F = k x increases linearly with x. Work done in stretching from 0 to x equals the area under the F vs x line, a triangle of area 1/2 * x * (k x) = 1/2 k x^2. This stored work is the elastic potential energy.
Yes. Because it contains x^2, the value is positive for both stretching (x > 0) and compression (x < 0). The energy is zero only at the natural length, x = 0.
Its SI unit is the joule (J). Since it is an energy, its dimensional formula is [M L^2 T^-2], the same as work and kinetic energy.
For a spring-block system with no friction, total mechanical energy 1/2 m v^2 + 1/2 k x^2 stays constant. At maximum compression or stretch, all kinetic energy becomes spring PE, which is the standard NEET numerical setup (like the car-and-spring example in NCERT).
