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Indian Navy Pilot Energy Transfer in Systems — practice questions

Explore 31 free Energy Transfer in Systems questions in Indian Navy Pilot Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Energy Transfer in Systems questions and exercises with answers

Showing questions 1–30 of 31.

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 →As per your syllabus, A · B equals which expression?Read the answer and worked solution →As per your syllabus, work done by a force perpendicular to the displacement (e.g. centripetal force in UCM) is which?Read the answer and worked solution →As per your syllabus, the work done by a VARIABLE force F(x) over displacement from x_i to x_f is which?Read the answer and worked solution →As per your syllabus, INSTANTANEOUS power is given by which expression in terms of F and v?Read the answer and worked solution →As per your syllabus, 1 horsepower (hp) is approximately equal to which value?Read the answer and worked solution →As per your syllabus, an ELASTIC collision conserves which two quantities?Read the answer and worked solution →Order the atomic models in this chapterRead the answer and worked solution →True or false: Rutherford's two difficultiesRead the answer and worked solution →Which model does this belong to?Read the answer and worked solution →Complete Bohr's first postulateRead the answer and worked solution →True or false: line spectraRead the answer and worked solution →Charge and neutralityRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Tap every problem Bohr's model was built to solveRead the answer and worked solution →Flip and pair: model with its pictureRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Strength or weakness?Read the answer and worked solution →Complete the size of the nucleusRead the answer and worked solution →Reading the models backRead the answer and worked solution →Bohr's postulatesRead the answer and worked solution →Explained, or left unexplained?Read the answer and worked solution →True or false: the nucleusRead the answer and worked solution →Order the argument: why Rutherford's atom should collapseRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Assertion and reason: modelsRead the answer and worked solution →Spot the one that does not belongRead 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:

  1. $kx$
  2. $\dfrac{1}{2} k x^2$
  3. $k x^2$
  4. $\dfrac{1}{2} k^2 x$
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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:

  1. The change in its kinetic energy
  2. The change in its potential energy
  3. The total mechanical energy
  4. The momentum of the body
Show answer and solution

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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As per your syllabus, A · B equals which expression?

  1. A B (no factor)
  2. A B cos θ
  3. A B sin θ
  4. A B tan θ
Show answer and solution

Correct answer: B. A B cos θ

Your syllabus: scalar (dot) product A · B = AB cos θ. Maximum at θ = 0; zero at θ = 90°.

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