JEE Advanced Unit 1A: Kinematics: 1D Motion, Position & Velocity — practice questions — page 2

Explore 32 free Unit 1A: Kinematics: 1D Motion, Position & Velocity questions in JEE Advanced Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 1A: Kinematics: 1D Motion, Position & Velocity questions and exercises with answers

Showing questions 31–32 of 32.

The elastic potential energy stored in a spring of spring constant $k$ when stretched by displacement $x$ is:Read the answer and worked solution →The work-energy theorem states that the net work done on a body equals:Read the answer and worked solution →

Worked practice questions

The elastic potential energy stored in a spring of spring constant $k$ when stretched by displacement $x$ is:

$kx$
$\dfrac{1}{2} k x^2$
$k x^2$
$\dfrac{1}{2} k^2 x$

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Correct answer: B. $\dfrac{1}{2} k x^2$

Force needed to stretch a spring by $x$ is $F = kx$ (Hooke's law). Work done in stretching from 0 to $x$: $U = \int_0^x kx' \, dx' = \dfrac{1}{2} k x^2$. The $\tfrac{1}{2}$ shows up because force grows linearly during the stretch — average force is $kx/2$, total work is $kx/2 \cdot x = \tfrac{1}{2} k x^2$. Option A ($kx$) is the *force*, not the energy.

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The work-energy theorem states that the net work done on a body equals:

The change in its kinetic energy
The change in its potential energy
The total mechanical energy
The momentum of the body

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Correct answer: A. The change in its kinetic energy

Work-energy theorem: $W_{\text{net}} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}}$. This is a direct consequence of Newton's second law. Apply $F = ma$ and integrate over displacement: $\int F \, dx = \int m a \, dx = \int m v \, dv = \tfrac{1}{2} m v^2 \big|_{u}^{v} = \Delta KE$. If only conservative forces act, this is equivalent to saying $\Delta KE + \Delta PE = 0$, the conservation of mechanical energy. With friction or other dissipative forces present, energy is lost to heat and the strict mechanical-energy conservation breaks, but the work-energy theorem still holds because work includes *all* forces.

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