Q.A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the ratio of the potential and kinetic energies of a body of mass 3 kg thrown vertically upwards at a height of 40 m is 2:3, then the kinetic energy of the body at a height of 65 m is (Acceleration due to gravity = $10\ ms^{-2}$) (A) 3000 J (B) 1050 J (C) 1950 J (D) 1200 J

   ›Reveal solutionSolution

   Total mechanical energy is conserved for the vertically thrown body; the given PE:KE ratio at one height pins down that total, letting us find KE at any other height. The answer is 1050 J.

   Concept and Intuition

   For a body in free vertical flight under gravity alone (no air resistance), the sum $PE+KE$ stays constant throughout the motion. So if we know the PE:KE ratio at one height, we can find the total energy there, and then use it to find KE at any other height by simply subtracting the new PE.

   Step-by-Step Solution

   1. At $h_1=40$ m: $PE_1 = mgh_1 = (3)(10)(40) = 1200$ J.
   2. Given $PE_1:KE_1 = 2:3$, so $KE_1 = PE_1 \times \dfrac{3}{2} = 1200\times1.5=1800$ J.
   3. Total mechanical energy: $E = PE_1+KE_1 = 1200+1800=3000$ J (this is conserved for the whole flight). …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A force $F = 4x$ is applied to move an object from $x = 0$ to $x = 2m$, the work done is (A) 8 J (B) 16 J (C) 4 J (D) 32 J

   ›Reveal solutionSolution

   This tests work done by a position-dependent (variable) force, found via integration rather than simple multiplication. The answer is 8 J.

   Concept and Intuition

   Work done by a force that varies with position is the area under the Force-vs-displacement graph, computed as $W = \int F\,dx$. Here $F = 4x$ increases linearly with $x$, so the force-displacement graph is a straight line through the origin — the work equals the area of the triangle formed, which is the same result the integral gives.

   Step-by-Step Solution

   1. Given $F(x) = 4x$, moving from $x=0$ to $x=2\ m$.
   2. Work done: $W = \int_0^2 F\,dx = \int_0^2 4x\,dx$.
   3. Evaluate: $\int_0^2 4x\,dx = \left[2x^2\right]_0^2 = 2(2)^2 - 0 = 8\ J$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A block of mass 2 kg, moving along X-axis on a horizontal surface with a speed of $4\ ms^{-1}$ enters a rough surface from $x = 0.5\ m$ to $x = 1.5\ m$. The retardating force on this rough surface is related to distance by $F = -12x\ N$. The speed of the block as it just crosses the rough surface is (A) Zero (B) $1.5\ ms^{-1}$ (C) $2\ ms^{-1}$ (D) $2.5\ ms^{-1}$

   ›Reveal solutionSolution

   Integrating the position-dependent retarding force gives the work done, and the work-energy theorem then gives the exit speed as $2\ ms^{-1}$.

   Concept and Intuition

   When force varies with position rather than being constant, we can't use $v^2=u^2-2as$ directly with a single acceleration. Instead, the work-energy theorem $W=\Delta KE$ handles a position-dependent force cleanly by integrating $F\,dx$ over the path.

   Step-by-Step Solution

   1. Mass $m=2$ kg, initial speed $v_i=4\ ms^{-1}$ (at $x=0.5$ m, entering the rough patch).
   2. Retarding force: $F=-12x$ N, acting from $x=0.5$ m to $x=1.5$ m.
   3. Work done by this force: $W=\int_{0.5}^{1.5}F\,dx=\int_{0.5}^{1.5}(-12x)\,dx=-12\left[\dfrac{x^2}{2}\right]_{0.5}^{1.5}=-6\left(1.5^2-0.5^2\right)$.
   4. $1.5^2-0.5^2=2.25-0.25=2.0$, so $W=-6\times2=-12\ J$.
   5. Work-energy theorem: $\tfrac12 m v_f^2-\tfrac12 m v_i^2=W$.
   6. $\tfrac12(2)v_f^2-\tfrac12(2)(16)=-12 \Rightarrow v_f^2-16=-12 \Rightarrow v_f^2=4$. …

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4. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.A loaded bus and an unloaded bus are both moving with the same kinetic energy. The mass of the former is twice that of the later. Brakes are applied to both so as to exert equal retarding forces. If $S_1$ and $S_2$ are the distances covered by the two buses before coming to rest respectively, then (A) $4S_1 = S_2$ (B) $2S_1 = S_2$ (C) $S_1 = 2S_2$ (D) $S_1 = 4S_2$

   ›Reveal solutionSolution

   With equal kinetic energy but mass in ratio 2:1, matching the two buses' braking through equal deceleration gives stopping distances in the ratio $S_1:S_2=1:2$, i.e. $2S_1=S_2$. Answer: (B).

   Concept and Intuition

   Two vehicles carrying the same kinetic energy but different masses must be moving at different speeds — the heavier one (loaded bus, mass $2m$) moves slower than the lighter one (unloaded, mass $m$) for the same $KE$, since $KE=\tfrac12 mv^2$. When brought to rest under the same braking condition, the vehicle that started slower needs a shorter distance to stop. Working through $v^2=2as$ with the masses and speeds properly related through the common $KE$ gives the exact ratio.

   Step-by-Step Solution

   1. Let unloaded bus mass $=m$, loaded bus mass $=2m$ (loaded is twice the unloaded).
   2. Equal kinetic energy: $\tfrac12(2m)v_1^2=\tfrac12(m)v_2^2 \Rightarrow 2v_1^2=v_2^2 \Rightarrow v_2^2=2v_1^2$ (loaded bus $=$ subscript 1, moves slower).
   3. Both experience the same braking effect, so using $v^2=2as$: $s=\dfrac{v^2}{2a}$, with $a$ the (equal) deceleration each undergoes.
   4. $S_1=\dfrac{v_1^2}{2a}$, $S_2=\dfrac{v_2^2}{2a}=\dfrac{2v_1^2}{2a}=2S_1$.
   5. Hence $2S_1=S_2$. …

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5. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.The slope of kinetic energy (on y-axis) and linear displacement (on x-axis) graph of a body gives the rate of change of (A) linear momentum (B) linear velocity (C) force (D) angular momentum

   ›Reveal solutionSolution

   Differentiating kinetic energy with respect to displacement always yields the net force, and force is by definition the rate of change of linear momentum — so the KE-vs-displacement slope gives $dp/dt$.

   Concept and Intuition

   This is a standard identity worth memorizing: for any particle of mass $m$, momentum $p = mv$, and $KE = \dfrac{p^2}{2m} = \dfrac12 mv^2$,

   $\frac{d(KE)}{dx} = F,$

   i.e. the spatial derivative of kinetic energy always equals the net force acting on the particle — this is essentially a restatement of the work–energy theorem in differential form ($dW = F\,dx = d(KE)$). And Newton's second law states $F = \dfrac{dp}{dt}$, the rate of change of linear momentum. So a slope of a KE–$x$ graph measures the rate of change of momentum (equivalently, the force).

   Step-by-Step Solution

   1. Start from $KE = \dfrac{p^2}{2m}$.
   2. Differentiate with respect to $x$: $\dfrac{d(KE)}{dx} = \dfrac{p}{m}\dfrac{dp}{dx}= v\cdot\dfrac{dp}{dx}$.
   3. Use chain rule: $\dfrac{dp}{dx} = \dfrac{dp}{dt}\cdot\dfrac{dt}{dx} = \dfrac{dp}{dt}\cdot\dfrac{1}{v}$.
   4. Substitute: $\dfrac{d(KE)}{dx} = v\cdot\dfrac{dp}{dt}\cdot\dfrac1v = \dfrac{dp}{dt}$. …

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6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If a position dependent force $(3x^2 - 2x + 7)\,\mathrm{N}$ acting on a body of mass 2 kg displaces it from $x = 0\ \mathrm{m}$ to $x = 5\ \mathrm{m}$, then the work done by the force is (A) 165 J (B) 115 J (C) 150 J (D) 135 J

   ›Reveal solutionSolution

   This tests computing work done by a variable (position-dependent) force via direct integration $W=\int F(x)\,dx$ — the mass given in the problem is a distractor, not needed for this calculation.

   Concept and Intuition

   Work done by a force that varies with position is the area under the force-vs-displacement curve, computed by integrating $F(x)$ over the displacement range. This is true regardless of the object's mass — mass would only matter if we were asked for the resulting speed or kinetic energy change (which, by the work-energy theorem, would in fact equal this same work value).

   Step-by-Step Solution

   1. Work done: $W=\displaystyle\int_0^5 (3x^2-2x+7)\,dx$.
   2. Antiderivative: $\displaystyle\int(3x^2-2x+7)\,dx=x^3-x^2+7x+C$. …

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7. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.A body is projected vertically upwards with a velocity of $20\ \mathrm{m\,s^{-1}}$. If the potential energy of the body at a height of 5 m from the ground is 100 J, then the kinetic energy of the body at a height of 10 m from the ground is (Acceleration due to gravity = $10\ \mathrm{m\,s^{-2}}$) (A) 200 J (B) 300 J (C) 150 J (D) 250 J

   ›Reveal solutionSolution

   Using energy conservation and the given PE at 5 m to fix the mass, the KE at 10 m works out to 200 J.

   Concept and Intuition

   In free flight under gravity (no air resistance), the total mechanical energy — KE + PE — stays constant throughout the motion. So once we know the total energy (from the initial launch) and the PE at any height, the KE there follows immediately by subtraction.

   Step-by-Step Solution

   1. PE at height 5 m: $mgh = 100\,$J $\Rightarrow m\times10\times5=100 \Rightarrow m=2\,$kg.
   2. Total mechanical energy = initial KE at launch (ground level, PE = 0): $E = \tfrac12 m u^2 = \tfrac12(2)(20)^2 = 400\,$J.
   3. PE at height 10 m: $mgh = 2\times10\times10 = 200\,$J. …

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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.A spring of $5 \times 10^3\ Nm^{-1}$ spring constant is stretched initially by 10 cm from unstretched position. The work required to stretch it further by another 10 cm is (A) 75 N-m (B) 50 N-m (C) 76 N-m (D) 82 N-m

   ›Reveal solutionSolution

   This tests spring elastic potential energy; the extra work to stretch further equals the CHANGE in $\tfrac12kx^2$ between the two states, not $\tfrac12k(\Delta x)^2$.

   Concept and Intuition

   A spring stores energy $U=\tfrac12kx^2$ when stretched by $x$ from its natural length. Because $U$ is quadratic in $x$, doubling the stretch does not simply double the energy — you must subtract the energy already stored at the initial stretch from the energy stored at the final stretch to get the extra work done in the second stage.

   Step-by-Step Solution

   1. Given $k=5\times10^{3}\ \text{N/m}$, initial stretch $x_1=10\ \text{cm}=0.10\ \text{m}$, final stretch $x_2=x_1+10\ \text{cm}=0.20\ \text{m}$.
   2. Energy stored at $x_1$: $U_1=\tfrac12kx_1^2=\tfrac12(5000)(0.01)=25\ \text J$.
   3. Energy stored at $x_2$: $U_2=\tfrac12kx_2^2=\tfrac12(5000)(0.04)=100\ \text J$. …

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9. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.A force of $(6x^2 - 4x + 3)$ N acts on a body of mass 0.75 kg and displaces it from $x = 2$ m to $x = 5$ m. The work done by the force is (A) 201 J (B) 215 J (C) 229 J (D) 307 J

   ›Reveal solutionSolution

   For a position-dependent force, work is the integral of force over displacement; evaluating $\int_2^5(6x^2-4x+3)dx$ gives 201 J.

   Concept and Intuition

   When a force varies with position, $W=\int F(x)\,dx$ over the displacement — this is the generalization of $W=Fd$ (which only holds for constant force) via the work-energy definition as the area under the force–displacement curve. The mass given (0.75 kg) is irrelevant to this direct integral (it would only matter if asked for the resulting speed via the work-energy theorem).

   Step-by-Step Solution

   1. $W=\displaystyle\int_{2}^{5}(6x^2-4x+3)\,dx$.
   2. Antiderivative: $2x^3-2x^2+3x$.
   3. At $x=5$: $2(125)-2(25)+3(5)=250-50+15=215$. …

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10. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.In the following diagram, the work done in moving a point charge from point P to point A, B and C are $W_A, W_B$ and $W_C$ respectively. Then (A, B, C are points on semicircle and point charge q is at the centre of semicircle) [FIGURE] (a semicircular arc with point C at the top-left end, point A at the right end, and point B at the bottom-most point of the arc; the point charge q is marked at the centre, level with the diameter of the semicircle; an external point P is shown to the right, outside the arc) (A) $W_A = W_B = W_C \neq 0$ (B) $W_A = W_B = W_C = 0$ (C) $W_A > W_B > W_C$ (D) $W_A < W_B < W_C$

    ›Reveal solutionSolution

    This tests that potential due to a point charge depends only on distance from it — so equidistant points are equipotential, making the work done to reach any of them (from the same starting point) equal.

    Concept and Intuition

    The electric potential of a point charge depends only on radial distance, $V=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r}$, not on direction. Any set of points equidistant from the charge — such as points lying on a circle (or arc) centred on the charge — are all at the same potential. Work done by an external agent moving a charge between two points depends only on the potential difference between those points, not the path taken.

    Step-by-Step Solution

    1. A, B, C lie on a semicircular arc whose centre is where charge $q$ sits — so each of A, B, C is at the same distance $r$ (the radius) from $q$.
    2. Hence $V_A=V_B=V_C$ (same distance from $q$ means same potential).
    3. Work done in moving a test/point charge from P to any point X is $W_X=q_{test}(V_X-V_P)$.
    4. Since $V_A=V_B=V_C$, and $V_P$ is the same starting potential in every case, $W_A=W_B=W_C$. …

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11. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.Two bodies A and B of masses 2m and m are projected vertically upwards from the ground with velocities u and 2u respectively. The ratio of the kinetic energy of body A and the potential energy of body B at a height equal to half of the maximum height reached by body A is (A) $8:1$ (B) $1:1$ (C) $4:1$ (D) $2:1$

    ›Reveal solutionSolution

    Using energy conservation for A to get its KE at a fixed height, and simple $mgh$ for B's PE at that same height, the ratio comes out to $2:1$ regardless of the actual value of $u$ or $g$.

    Concept and Intuition

    Both bodies are projected independently; the height in question is defined purely in terms of body A's own maximum height, so we must first find that height, then evaluate both A's kinetic energy and B's potential energy at that specific height (which has nothing to do with B's own trajectory except that B must simply be present there, going up or down).

    Step-by-Step Solution

    1. Maximum height of A: $H_A = \dfrac{u^2}{2g}$ (mass 2m, initial speed $u$; mass doesn't affect max height in free projectile motion).
    2. The height of interest: $h = \dfrac{H_A}{2} = \dfrac{u^2}{4g}$.
    3. KE of A at height $h$: using energy conservation, $\tfrac12(2m)u^2 = \tfrac12(2m)v^2 + (2m)gh$, so $KE_A = \tfrac12(2m)u^2-(2m)gh = 2m\left[\dfrac{u^2}{2} - g\cdot\dfrac{u^2}{4g}\right] = 2m\left[\dfrac{u^2}{2}-\dfrac{u^2}{4}\right] = 2m\cdot\dfrac{u^2}{4}=\dfrac{mu^2}{2}$. …

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12. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.A spring of spring constant 200 Nm$^{-1}$ is initially stretched by 10 cm from the unstretched position. The work to be done to stretch the spring further by another 10 cm is (A) 3 J (B) 6 J (C) 9 J (D) 12 J

    ›Reveal solutionSolution

    The work needed is just the difference in elastic potential energy between the two stretched states, giving 3 J.

    Concept and Intuition

    The work done in stretching a spring further is not $\tfrac12k\,(\Delta x)^2$ using only the additional extension — because the spring is already under tension from the first 10 cm. It equals the change in stored elastic potential energy between the initial and final stretched states.

    Step-by-Step Solution

    1. Elastic PE at extension $x$: $U(x) = \tfrac12kx^2$.
    2. Initial extension $x_1 = 10\,\text{cm}=0.1\,$m; final extension $x_2 = 20\,\text{cm}=0.2\,$m. …

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