Q.Develop the notions of work and kinetic energy and show that it leads to work-energy theorem. A machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600 ms^-1. If the mass of each bullet is 5 gm, find the power of the machine gun.

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the column - select the correct definition (from Column B) for the term 'Energy' (Column A):

   (a) change in linear momentum

   (b) motion opposing force

   (c) loss of energy

   (d) rate of change of momentum

   (e) ability of doing work

   (f) rate of doing work

   ›Reveal solutionSolution

   Energy is defined as the capacity of a body to do work.

   In mechanics, the energy possessed by a body (whether kinetic, due to its motion, or potential, due to its position/configuration) is precisely defined as its capacity to perform work — a body with more energy can do more work before that capacity is exhausted. Thi …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.A particle moves under the effect of a force F = αx from x = 0 to x = d. The work done in the process is (α = constant)

   (a) αd

   (b) αd^2

   (c) (1/2)αd^2

   (d) zero

   ›Reveal solutionSolution

   W = integral of alpha x dx from 0 to d = ½ alpha d^2. Answer (C).

   For a position-dependent force, work done = integral of F dx.

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.A force of 40 N acts on body of mass 5 kg which is initially at rest. What is the amount of work done in the first 10 s?

   (a) 1600 J

   (b) -1600 J

   (c) 400 J

   (d) -400 J

   ›Reveal solutionSolution

   Work done by a constant force on a body starting from rest is $W = F^2t^2/(2m)$. With the numbers as printed the arithmetic gives 16{,}000 J (ten times option (a)); using $m = 50$ kg — the value consistent with the printed options, and a very plausible OCR/typo of '5 kg' for '50 kg' in the source paper — gives exactly 1600 J, so that is the answer selected here.

   Acceleration: $a = F/m$

   Distance covered from rest in time $t$: $s = (1/2)at^2$

   Work done by the applied force: $W = F \times s = F \times (1/2)(F/m)t^2 = F^2t^2/(2m)$

   With $F = 40$ N, $t = 10$ s, and $m = 50$ kg (see note above):

   $a = 40/50 = 0.8$ m/s²

   $s = (1/2)(0.8)(10)^2 = 40$ m

   $W = 40 \times 40 = 1600$ J

   …

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4. CBSE 2024Set ANNUAL1 markMCQ
   Q.The dimensional formula of work done is the same as the dimensional formula of

   (a) Momentum

   (b) Power

   (c) Energy

   (d) Torque

   ›Reveal solutionSolution

   Work and energy are dimensionally identical because work done equals the change in energy (work-energy theorem).

   Work done, W = Force x displacement = [MLT^-2][L] = [M L^2 T^-2].

   Energy (kinetic or potential) has the same dimensional formula [M L^2 T^-2], since energy is literally defined as the capacity to do work.

   Check the other options:

   • Momentum = mass x velocity = [M][LT^-1] = [MLT^-1] (different).
   • Power = work/time = [ML^2T^-3] (different). …

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5. CBSE 2024Set ANNUAL1 markMCQ
   Q.Energy is

   (a) capacity of doing work

   (b) rate of doing work

   (c) change of work

   (d) none of these

   ›Reveal solutionSolution

   Energy is defined as the capacity of a body or system to do work.

   By definition, energy is the ability of a system to perform work; a body with more energy can do more work. This is why work and energy share the same SI unit, the joule, and the same dimensional fo …

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6. CBSE 2024Set ANNUAL1 markMCQ
   Q.Area under force-displacement curve represents

   (a) Velocity

   (b) Acceleration

   (c) Impulse

   (d) Work done

   ›Reveal solutionSolution

   Work done = force x displacement (for a varying force, the integral of F dx), which is exactly the area enclosed under an F-x graph.

   For a constant force, W = F x s, which is literally the area of the rectangle under a horizontal F-x line. For a varying force, W = ∫ F dx, the area under the (possibly curved) F-x graph, found by summi …

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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.Which of the following is a unit of energy?

   (a) Horse power

   (b) Joule second

   (c) Kilowatt hour

   (d) Watt.

   ›Reveal solutionSolution

   Kilowatt hour = power × time = energy, unlike horse power and watt (power) or joule-second (action).

   …

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8. CBSE 2024Set ANNUAL1 mark
   Q.State whether true or false: Work is a vector quantity.

   ›Reveal solutionSolution

   False. Work W = F.d (dot product of force and displacement) is a scalar quantity.

   Work done by a force F causing a displacement d is defined as W = F . d = |F||d| cos(theta), the scalar (dot) product of the two vectors. …

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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.The work done by a body against friction always results in

   (1) loss of kinetic energy

   (2) loss of potential energy

   (3) gain of kinetic energy

   (4) gain of potential energy

   ›Reveal solutionSolution

   Friction is dissipative: work done against it always drains kinetic energy from the system, converting it irreversibly into heat.

   Friction opposes relative motion, so when a body moves against friction, the friction force does negative work on the body. By the work-energy theorem, negative work done on a body reduces its kinetic energy. This lost mechanical energy is not stored as potential energy (friction is not a conservative force, so no potential energy can be associated with it) - it is dissipated as heat at the su …

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10. CBSE 2023Set ANNUAL1 markMCQ
    Q.A raindrop of mass 1 g falling from a height of 1 km hits the ground with a speed of 50 ms^-1. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take g = 10 ms^-2)

    (1) 10 J

    (2) -10 J

    (3) 8.75 J

    (4) -8.75 J

    ›Reveal solutionSolution

    Apply the work-energy theorem: total work done (by gravity and the resistive force together) equals the change in kinetic energy; solve for the unknown resistive-force work.

    Given:

    Mass, m = 1 g = 0.001 kg

    Height fallen, h = 1 km = 1000 m

    Final speed, v = 50 m/s

    g = 10 m/s^2

    Work done BY gravity (positive, since displacement is in the direction of gravity):

    W_gravity = mgh = 0.001 x 10 x 1000 = 10 J

    Change in kinetic energy (drop starts from rest):

    ΔKE = (1/2)mv^2 - 0 = (1/2)(0.001)(50)^2 = (1/2)(0.001)(2500) = 1.25 J

    …

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11. CBSE 2023Set ANNUAL1 mark
    Q.Mention two physical quantities which have same dimensions as that of work.

    ›Reveal solutionSolution

    Work has dimensional formula $[M L^2 T^{-2}]$; energy and torque share exactly this formula.

    Work is defined as $W = F\cdot d$, giving dimensions $[MLT^{-2}][L] = [M L^2 T^{-2}]$. By the work-energy theorem, energy (kinetic energy $\tfrac12mv^2$, potential energy $mgh$, heat, etc.) has exactly the same dimensional formula, since energy is measured in the same unit, the joule. Torque is defined as $\tau = F\times r$ (force times perpendicular distance), which also gives $[MLT^{-2}][L] = [ML^2T^{-2}]$ — the same d …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.Which of the following quantities represents the change in kinetic energy of any object?

    (a) Force

    (b) Mass

    (c) Linear momentum

    (d) Work

    ›Reveal solutionSolution

    The work-energy theorem directly identifies work as the physical quantity equal to the change in kinetic energy of a body.

    Starting from Newton's second law, F = ma = m(dv/dt). The work done by this force over a displacement ds is:

    dW = F ds = m (dv/dt) ds = m v dv (since ds/dt = v)

    Integrating from initial speed u to final speed v:

    W = integral of m v dv from u to v = (1/2) m v^2 - (1/2) m u^2 = KE_final - KE_initial = Delta KE

    …

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