Q.Consider Example 5.8 taking the coefficient of friction, $\mu$, to be $0.5$ and calculate the maximum compression of the spring.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The energy required to transfer a satellite of mass ‘m’ from an orbit of height 0.5R from the surface of the earth to an orbit of height 2R from the surface of the earth is (g - acceleration due to gravity on the surface of the earth and R - radius of the earth) (A) $\dfrac{mgR}{4}$ (B) $\dfrac{mgR}{2}$ (C) $\dfrac{mgR}{6}$ (D) $\dfrac{mgR}{3}$

   ›Reveal solutionSolution

   The energy required is the difference in total mechanical energy between the two orbits. For a satellite in a circular orbit, total energy = $-\frac{GMm}{2r}$. Using $g = GM/R^2$, the answer simplifies to $\frac{mgR}{6}$.

   The key idea here is that a satellite in a circular orbit has a total mechanical energy that is negative and exactly half its gravitational potential energy. To move from one orbit to another, you must supply the difference in these total energies — that’s the work needed, regardless of the path taken.

   Why half? Because the satellite’s kinetic energy is positive and exactly half the magnitude of its negative potential energy, giving a neat cancellation. This is a direct consequence of the centripetal force condition for a circular orbit.

   Let’s work it through.

   1. Write the total energy for a circular orbit. For a satellite of mass $m$ orbiting Earth (mass $M$) at a distance $r$ from the centre, the gravitational force provides the centripetal force:

   $\frac{GMm}{r^2} = \frac{mv^2}{r}$

   So kinetic energy $K = \frac12 mv^2 = \frac{GMm}{2r}$.

   Potential energy $U = -\frac{GMm}{r}$.

   Total mechanical energy:

   $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

   This is a central formula to remember.

   $E_{\text{orbit}} = -\frac{GMm}{2r}$

   2. Find the orbital radii from the given heights.

      Height is measured from Earth’s surface. Radius of Earth = $R$.

      • For height $0.5R$: orbital radius $r_1 = R + 0.5R = 1.5R = \frac{3R}{2}$
      • For height $2R$: orbital radius $r_2 = R + 2R = 3R$

   3. Compute the total energy in each orbit.

   $E_1 = -\frac{GMm}{2r_1} = -\frac{GMm}{2 \cdot \frac{3R}{2}} = -\frac{GMm}{3R}$

   $E_2 = -\frac{GMm}{2r_2} = -\frac{GMm}{2 \cdot 3R} = -\frac{GMm}{6R}$

   4. Energy required = difference in total energy. …

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2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the work done to double the velocity of a body from $15\ \mathrm{ms}^{-1}$ is K times the work done to double its velocity from $10\ \mathrm{ms}^{-1}$, then the value of K is (A) 2.75 (B) 2.25 (C) 3.25 (D) 3.75

   ›Reveal solutionSolution

   The work done equals the change in kinetic energy. Doubling the velocity from $15\ \mathrm{m/s}$ gives a larger kinetic-energy increase than from $10\ \mathrm{m/s}$, and the ratio $K$ is found to be $2.25$, corresponding to option (B).

   Concept & Intuition

   Work–energy theorem: the work done on a body equals its change in kinetic energy.

   Kinetic energy is $KE = \frac12 m v^2$.

   When we “double the velocity” from an initial speed $u$ to $2u$, the change in kinetic energy is

   $\Delta KE = \frac12 m (2u)^2 - \frac12 m u^2 = \frac12 m (4u^2 - u^2) = \frac12 m (3u^2).$

   So the work needed is proportional to $u^2$. The problem asks for the ratio of work when $u=15$ to work when $u=10$ — that ratio is simply $(15^2)/(10^2) = 225/100 = 2.25$. No mass needed; it cancels.

   Step-by-step reasoning

   1. Work done = change in kinetic energy For a body of mass $m$, the work $W$ to change its speed from $v_i$ to $v_f$ is

   $W = \frac12 m v_f^2 - \frac12 m v_i^2.$

   2. Case 1: doubling from $10\ \mathrm{m/s}$ Initial speed $u_1 = 10$, final speed $2u_1 = 20$.

   $W_1 = \frac12 m (20^2 - 10^2) = \frac12 m (400 - 100) = \frac12 m (300).$

   3. Case 2: doubling from $15\ \mathrm{m/s}$ Initial speed $u_2 = 15$, final speed $2u_2 = 30$.

   $W_2 = \frac12 m (30^2 - 15^2) = \frac12 m (900 - 225) = \frac12 m (675).$

   4. Find $K = W_2 / W_1$ $$$$

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.To drive a vertical nail of mass 10 g in to wood through 9 cm, an iron block of mass 990 g is dropped on to it freely from a height of 10 m above the nail. If the collision between the nail and the block is perfectly inelastic, then the force of resistance offered by wood is (Acceleration due to gravity $= 10\,\mathrm{ms}^{-2}$) (A) 898 N (B) 989 N (C) 1089 N (D) 1198 N

   ›Reveal solutionSolution

   The block reaches $\sqrt{200}\ \text{m s}^{-1}$, the inelastic collision leaves $\approx 98\ \text{J}$ of kinetic energy in the nail+block, and dissipating it over $9\ \text{cm}$ needs a resistance of $\approx 1089\ \text{N}$.

   1. Speed of the block on reaching the nail (free fall through $h=10\ \text{m}$):

   $v=\sqrt{2gh}=\sqrt{2(10)(10)}=\sqrt{200}=14.14\ \text{m s}^{-1}.$

   2. Perfectly inelastic collision ($M=0.99\ \text{kg}$ block, $m=0.01\ \text{kg}$ nail). Conserving momentum:

   $V=\frac{Mv}{M+m}=\frac{0.99\sqrt{200}}{1.0}=14.0\ \text{m s}^{-1}.$

   3. Kinetic energy of the combined mass just after impact: …

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4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Two identical rain drops are falling through air each with a terminal velocity ‘V’. If the two drops coalesce to form a single big drop, then the terminal velocity of the big drop is (A) $\dfrac{V}{2^{2/3}}$ (B) $2^{2/3} V$ (C) $2^{1/3} V$ (D) $\dfrac{V}{2^{1/3}}$

   ›Reveal solutionSolution

   When a body falls through a viscous medium, it reaches a constant terminal velocity when the gravitational force is balanced by the buoyant force and the viscous drag. For a spherical drop, this terminal velocity is proportional to the square of its radius. When two identical drops coalesce, the total volume is conserved, leading to a larger radius for the new drop. This larger radius results in a higher terminal velocity, specifically $2^{2/3}$ times the original velocity. The final answer is $\boxed{2^{2/3} V}$.

   The problem asks us to find the terminal velocity of a large drop formed by the coalescence of two identical smaller raindrops, each falling with a terminal velocity $V$. To solve this, we need to understand what terminal velocity is and how it depends on the physical properties of the drop, especially its size.

   Concept: Terminal Velocity

   When an object falls through a fluid (like air), it experiences three main forces:

   1. Gravitational Force ($F_g$): Acting downwards, due to the object's mass.
   2. Buoyant Force ($F_b$): Acting upwards, due to the displacement of the fluid.
   3. Viscous Drag Force ($F_v$): Acting upwards, opposing the motion, due to the fluid's resistance.

   Initially, as the object starts falling, its speed increases, and so does the viscous drag force. Eventually, the upward forces (buoyant force + viscous drag) become equal to the downward gravitational force. At this point, the net force on the object is zero, and it stops accelerating, continuing to fall at a constant maximum velocity called the terminal velocity ($v_t$).

   For a small spherical object of radius $r$ falling through a fluid of viscosity $\eta$, the viscous drag force is given by Stokes' Law:

   $F_v = 6\pi \eta r v_t$

   The gravitational force on the drop is $F_g = mg$, where $m$ is the mass of the drop. If $\rho_w$ is the density of water (the drop) and $V_d$ is its volume, then $m = \rho_w V_d$. For a sphere, $V_d = \frac{4}{3}\pi r^3$. So, $F_g = \frac{4}{3}\pi r^3 \rho_w g$.

   The buoyant force is $F_b = \rho_a V_d g$, where $\rho_a$ is the density of air (the fluid). So, $F_b = \frac{4}{3}\pi r^3 \rho_a g$.

   At terminal velocity, the forces balance:

   $F_g = F_b + F_v$

   $\frac{4}{3}\pi r^3 \rho_w g = \frac{4}{3}\pi r^3 \rho_a g + 6\pi \eta r v_t$

   Rearranging to solve for $v_t$:

   $6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho_w - \rho_a) g$

   $v_t = \frac{4\pi r^3 (\rho_w - \rho_a) g}{3 \cdot 6\pi \eta r}$

   $v_t = \frac{2 r^2 (\rho_w - \rho_a) g}{9 \eta}$

   The terminal velocity $v_t$ of a spherical drop of radius $r$ is given by:

   $v_t = \frac{2 r^2 (\rho_w - \rho_a) g}{9 \eta}$

   where $\rho_w$ is the density of the drop, $\rho_a$ is the density of the fluid (air), $g$ is the acceleration due to gravity, and $\eta$ is the coefficient of viscosity of the fluid.

   From this formula, we can see that for a given fluid and drop material, the terminal velocity is directly proportional to the square of the drop's radius:

   $v_t \propto r^2$

   This relationship is key to solving the problem.

   ---

   Here's how we solve the problem step-by-step:

   1. Express the terminal velocity of a single small drop:

      Let $r$ be the radius of each small raindrop. The terminal velocity of each small drop is given as $V$.

      From the formula derived above, we can write:

      $V = k r^2$

      where $k = \frac{2 (\rho_w - \rho_a) g}{9 \eta}$ is a constant that depends on the properties of water, air, and gravity, but not on the size of the drop.

   2. Determine the radius of the big drop after coalescence:

      When two identical raindrops coalesce, their total volume is conserved. …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Due to the presence of air resistance, if a body dropped from a height of 20 m reaches the ground with a speed of $18\,\mathrm{ms}^{-1}$, then the time taken by the body to reach the ground is nearly (A) $1.8\,\mathrm{s}$ (B) $2.2\,\mathrm{s}$ (C) $2\,\mathrm{s}$ (D) $2.5\,\mathrm{s}$

   ›Reveal solutionSolution

   The key idea is that with air resistance, the motion is not uniformly accelerated, so we cannot use constant‑acceleration formulas directly. Instead, we use the average velocity concept: for any motion, distance = average velocity × time. The average velocity is (initial speed + final speed)/2 only if acceleration is constant, but here we must approximate because the problem gives no details about the drag law. Using the given data, the simplest reasonable estimate gives time ≈ 2.2 s, which matches option (B).

   ---

   Concept and intuition

   When a body falls under gravity with air resistance, its acceleration is not constant — it starts at $g$ and decreases as speed increases, eventually approaching zero if the fall is long enough. That means the usual kinematic equations ($s = ut + \frac12 at^2$, $v^2 = u^2 + 2as$) do not apply.

   However, we still have the fundamental definition:

   $\text{average velocity} = \frac{\text{total displacement}}{\text{total time}}.$

   If we can estimate the average velocity, we can find the time. For a fall from rest ($u = 0$) with a known final speed $v = 18\ \mathrm{m/s}$, the average velocity lies somewhere between $0$ and $18$. In constant acceleration it would be exactly $(0+18)/2 = 9\ \mathrm{m/s}$. With air resistance, the body spends more time at lower speeds early on, so the average velocity is less than 9 m/s. That means the time will be greater than the constant‑acceleration time. Let’s quantify.

   ---

   Step‑by‑step reasoning

   1. Constant‑acceleration baseline (no air resistance) If there were no air resistance, the fall from 20 m with $u=0$ would satisfy

   $v^2 = 2gh \quad\Rightarrow\quad v = \sqrt{2 \times 9.8 \times 20} \approx \sqrt{392} \approx 19.8\ \mathrm{m/s}.$

   The actual final speed is only 18 m/s, so air resistance has reduced it. The time without resistance would be

   $t_{\text{no drag}} = \frac{v}{g} = \frac{19.8}{9.8} \approx 2.02\ \mathrm{s}.$

   This is close to option (C) 2 s, but that’s the no‑drag case — not our answer.

   2. Using average velocity with the given data For any motion,

   $t = \frac{s}{v_{\text{avg}}}.$

   We know $s = 20\ \mathrm{m}$. We need $v_{\text{avg}}$.

   With air resistance, the speed‑time graph is concave down (increasing but at a decreasing rate). The average value is less than the arithmetic mean of initial and final speeds. A common approximation (used when the exact drag law is unknown) is to take

   $v_{\text{avg}} \approx \frac{v_f}{2} \quad\text{or}\quad v_{\text{avg}} \approx \frac{v_f}{k}$

   with $k > 2$. But we can do better: we know the final speed is 18 m/s, and the initial is 0. For a body that starts from rest and experiences a resistive force proportional to speed (linear drag), the average velocity is exactly

   $v_{\text{avg}} = \frac{v_f}{\ln\!\left(\frac{v_f}{v_f - gt}\right)} \quad\text{(not simple)}.$

   However, the problem expects a quick estimate. Notice that the constant‑acceleration time (2.02 s) gives a final speed of ~19.8 m/s, but we have only 18 m/s. The reduction in speed is modest, so the time should be only slightly larger than 2.02 s. Among the options, 2.2 s is the next reasonable value.

   3. A more precise estimate using the work‑energy idea The work done by air resistance equals the loss in kinetic energy compared to the no‑drag case: $\text{Loss in KE} = \frac12 m (19.8^2 - 18^2) \approx \frac12 m (392 - 324) = 34m.$ …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Due to the presence of air resistance, if a body dropped from a height of $20\,\mathrm{m}$ reaches the ground with a speed of $18\,\mathrm{ms}^{-1}$, then the time taken by the body to reach the ground is nearly (A) $2.2\,\mathrm{s}$ (B) $2.5\,\mathrm{s}$ (C) $1.8\,\mathrm{s}$ (D) $2\,\mathrm{s}$

   ›Reveal solutionSolution

   The key idea is to treat the motion under constant air resistance as having a constant effective acceleration, found from the work–energy relation, then use that acceleration to compute the time. The computed time is approximately 2.2 s, so option (A) is correct.

   ---

   Concept and intuition

   In free fall without air resistance, an object dropped from 20 m would hit the ground at about $\sqrt{2gh} \approx \sqrt{2 \cdot 9.8 \cdot 20} \approx 19.8\ \mathrm{m/s}$. Here the final speed is only 18 m/s, so air resistance has done negative work, reducing the kinetic energy.

   If we assume the resistive force is constant (a common simplification for such problems), then the net downward acceleration is constant but smaller than $g$. We can find that effective acceleration from the work–energy theorem, then use kinematics to get the time.

   ---

   Step‑by‑step solution

   1. Set up the work–energy relation The work done by gravity is $mgh$. The work done by air resistance is $-F_{\text{air}} h$ (negative because it opposes motion). The net work equals the change in kinetic energy:

   $mgh - F_{\text{air}} h = \frac{1}{2} m v^2 - 0$

   where $h = 20\ \mathrm{m}$, $v = 18\ \mathrm{m/s}$, and $g = 9.8\ \mathrm{m/s^2}$.

   2. Solve for the net acceleration Divide through by $m$:

   $gh - \frac{F_{\text{air}}}{m} h = \frac{1}{2} v^2$

   The net downward acceleration is $a = g - \frac{F_{\text{air}}}{m}$. So:

   $a h = \frac{1}{2} v^2 \quad\Rightarrow\quad a = \frac{v^2}{2h}$

   Plug numbers:

   $a = \frac{18^2}{2 \cdot 20} = \frac{324}{40} = 8.1\ \mathrm{m/s^2}$

   This is the constant effective acceleration.

   3. Use kinematics to find time For constant acceleration from rest:

   $h = \frac{1}{2} a t^2 \quad\Rightarrow\quad t = \sqrt{\frac{2h}{a}}$

   Substitute:

   $$$$

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7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.A block of mass ‘m’ with an initial kinetic energy ‘E’ moves up an inclined plane of inclination ‘θ’. If ‘μ’ is the coefficient of friction between the plane and the body, the work done against friction before coming to rest is (A) $\mu E \cos \theta$ (B) $\dfrac{\mu E \cos \theta}{\sin \theta - \mu \cos \theta}$ (C) $\dfrac{E \mu \cos \theta}{\cos \theta + \sin \theta}$ (D) $\dfrac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta}$

   ›Reveal solutionSolution

   The work done against friction equals the initial kinetic energy times the ratio of the frictional force to the net retarding force along the incline. The correct expression is $\dfrac{\mu E \cos\theta}{\sin\theta + \mu\cos\theta}$, which corresponds to option (D).

   ---

   Concept & Intuition

   When a block slides up an incline, two forces oppose its motion: the component of gravity down the plane ($mg\sin\theta$) and kinetic friction ($\mu mg\cos\theta$). Both do negative work, draining the block’s initial kinetic energy $E$ until it stops. The work done against friction is just the part of that energy loss due to friction alone. Since both forces act over the same distance $d$, the fraction of $E$ that goes into friction equals the ratio of the frictional force to the total retarding force.

   ---

   Step-by-step reasoning

   1. Identify the forces doing work As the block moves up the incline, the net force opposing motion is

   $F_{\text{net}} = mg\sin\theta + \mu mg\cos\theta.$

   The first term is gravity’s component down the plane; the second is kinetic friction. Both act opposite to the displacement.

   2. Relate work and distance The total work done by these opposing forces equals the change in kinetic energy (from $E$ to $0$):

   $(mg\sin\theta + \mu mg\cos\theta)\, d = E,$

   where $d$ is the distance traveled along the incline before stopping.

   Solve for $d$:

   $d = \frac{E}{mg(\sin\theta + \mu\cos\theta)}.$

   3. Work done against friction Friction alone does work $W_f = (\mu mg\cos\theta)\, d$ (the force times distance). Substitute $d$:

   $W_f = \mu mg\cos\theta \cdot \frac{E}{mg(\sin\theta + \mu\cos\theta)} = \frac{\mu E \cos\theta}{\sin\theta + \mu\cos\theta}.$

   4. Interpret the result …

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8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.A body thrown vertically upwards from the ground reaches a maximum height ‘h’. The ratio of the kinetic and potential energies of the body at a height 40% of h from the ground is (A) $2:3$ (B) $3:2$ (C) $1:1$ (D) $4:9$

   ›Reveal solutionSolution

   Using conservation of mechanical energy, at 40% of the maximum height the kinetic energy is 60% of the initial energy, so the ratio KE:PE = 60:40 = 3:2, making the correct option (B).

   Concept and Intuition

   When a body is thrown vertically upward, its total mechanical energy (kinetic + potential) remains constant if we ignore air resistance. At the ground, all energy is kinetic. At the maximum height $h$, all energy is potential. At any intermediate height, the sum is the same. So if we know the fraction of the height, we know the fraction of potential energy, and the rest must be kinetic. The ratio follows directly.

   Step-by-step solution

   1. Define the total energy Let the mass of the body be $m$ and the initial speed at ground be $u$. At the ground, potential energy is zero, so total mechanical energy $E = \frac{1}{2} m u^2$. At the maximum height $h$, the speed is zero, so $E = mgh$. Hence

   $\frac{1}{2} m u^2 = mgh.$

   2. Energy at height $0.4h$ At a height $y = 0.4h$, the potential energy is

   $PE = mgy = mg(0.4h) = 0.4 \, mgh.$

   Since total energy $E = mgh$, the kinetic energy at that height is

   $KE = E - PE = mgh - 0.4\,mgh = 0.6\,mgh.$

   3. Find the ratio $$$$

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9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.A constant force of $(8\vec{i} - 2\vec{j} + 6\vec{k})$ N acting on a body of mass 2 kg displaces the body from $(2\vec{i} + 3\vec{j} - 4\vec{k})$ m to $(4\vec{i} - 3\vec{j} + 6\vec{k})$ m. The work done in the process is (A) $72$ J (B) $88$ J (C) $44$ J (D) $36$ J

   ›Reveal solutionSolution

   Work done by a constant force is the dot product of the force vector and the displacement vector. Here, the displacement is $(2\vec{i} - 6\vec{j} + 10\vec{k})$ m, and the dot product with $(8\vec{i} - 2\vec{j} + 6\vec{k})$ N gives $88$ J. The correct option is (B).

   Concept & Intuition

   Work done by a constant force is defined as the scalar product (dot product) of the force vector and the displacement vector. This is because only the component of force in the direction of displacement actually does work. The dot product automatically picks out that component. So we don’t need to worry about angles or components separately — just compute $\vec{F} \cdot \vec{d}$.

   Step-by-step solution

   1. Find the displacement vector Displacement $\vec{d}$ = final position − initial position. Initial position: $\vec{r}_i = 2\vec{i} + 3\vec{j} - 4\vec{k}$ Final position: $\vec{r}_f = 4\vec{i} - 3\vec{j} + 6\vec{k}$ So

   $\vec{d} = (4-2)\vec{i} + (-3-3)\vec{j} + (6 - (-4))\vec{k} = 2\vec{i} - 6\vec{j} + 10\vec{k}$

   2. Recall the work formula

      For a constant force $\vec{F}$, work done $W = \vec{F} \cdot \vec{d}$.

   3. Compute the dot product …

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10. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.A body of mass $3\,\text{kg}$ is moving under the action of a force which causes a displacement of $\left(\dfrac{t^3}{3}\right)\,\text{m}$, where ‘t’ is time in seconds. The work done by the force in first 2 seconds is (A) $2\,\text{J}$ (B) $3.8\,\text{J}$ (C) $5.2\,\text{J}$ (D) $24\,\text{J}$

    ›Reveal solutionSolution

    Work done equals the change in kinetic energy. By finding velocity from the given displacement, then computing kinetic energy at t = 2 s, we get 24 J. The correct option is (D).

    Concept & Intuition

    The problem gives displacement as a function of time: $s(t) = \frac{t^3}{3}$. Work done by a force is the integral of force over displacement, but a far simpler route uses the work–energy theorem: the net work done equals the change in kinetic energy. Since the body starts from rest (implied at t = 0, displacement is zero), the work done in the first 2 seconds is simply the kinetic energy at t = 2 s. We just need the velocity at that instant.

    Step-by-step solution

    1. Find velocity as a function of time Velocity is the derivative of displacement with respect to time:

    $v(t) = \frac{ds}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2$

    So at any time $t$, the speed is $v = t^2$ m/s.

    2. Compute velocity at t = 2 s

    $v(2) = (2)^2 = 4\ \text{m/s}$

    3. Apply the work–energy theorem Work done = change in kinetic energy. Initial kinetic energy (at t = 0) is zero because $v(0)=0$. $W = \Delta KE = \frac{1}{2} m v^2 - 0 = \frac{1}{2} \times 3 \times (4)^2$ …

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11. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.An engine is dragging a mass of $5000\,\mathrm{kg}$ with a velocity of $5\,\mathrm{ms}^{-1}$ along a smooth inclined plane of inclination $1$ in $50$. Then the power of the engine is (A) $5\,\mathrm{kW}$ (B) $2.5\,\mathrm{kW}$ (C) $10\,\mathrm{kW}$ (D) $25\,\mathrm{kW}$

    ›Reveal solutionSolution

    The engine must supply power to overcome the gravitational component along the incline at constant speed. The power is $P = mgv\sin\theta$, giving $P = 5000 \times 9.8 \times 5 \times \frac{1}{50} = 4900\,\mathrm{W} \approx 5\,\mathrm{kW}$, so the correct option is (A).

    Concept & Intuition

    When an object moves at constant velocity along a smooth (frictionless) incline, the only force the engine must overcome is the component of gravity pulling it back down the slope. Power is the rate at which work is done against this force. Since the plane is smooth, there is no friction to worry about — the engine’s entire output goes into lifting the mass against gravity at a steady rate.

    Step-by-step reasoning

    1. Identify the slope angle

       The incline is given as “1 in 50”, meaning for every 50 metres along the slope, the vertical rise is 1 metre. Hence $\sin\theta = \frac{1}{50}$.

    2. Force the engine must exert

       The gravitational force down the incline is $mg\sin\theta$. To keep the mass moving at constant velocity (no acceleration), the engine must pull with an equal and opposite force:

    $F = mg\sin\theta = 5000 \times 9.8 \times \frac{1}{50}$

    (We use $g = 9.8\,\mathrm{m/s^2}$ unless stated otherwise.)

    3. Calculate that force

    $F = 5000 \times 9.8 \times 0.02 = 5000 \times 0.196 = 980\,\mathrm{N}$

    4. Power formula For a constant force in the direction of motion, power is $P = Fv$, where $v$ is the speed. …

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12. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.A rocket moves straight upward with zero initial velocity and with an acceleration $20\ \mathrm{m/s^2}$. It runs out of fuel and stops accelerating at the end of $5^{\text{th}}$ sec. It reaches a maximum height and falls back to the earth. The speed when it hits the ground is (Take $g = 10\ \mathrm{m/s^2}$) (A) $100\sqrt{2}\ \mathrm{m/s}$ (B) $150\sqrt{3}\ \mathrm{m/s}$ (C) $50\sqrt{6}\ \mathrm{m/s}$ (D) $75\ \mathrm{m/s}$

    ›Reveal solutionSolution

    The rocket's motion is divided into three phases: initial acceleration, upward motion under gravity, and free fall. We calculate the velocity and height at the end of each phase to find the final speed when it hits the ground. The speed when it hits the ground is $\boxed{50\sqrt{6}\ \mathrm{m/s}}$.

    The problem describes the motion of a rocket in three distinct phases, each with a different constant acceleration. To solve this, we need to analyze each phase separately using the equations of motion for constant acceleration. The key idea is that the final velocity and position of one phase become the initial velocity and position for the next phase.

    Here's the breakdown of the rocket's journey:

    1. Phase 1: Accelerated motion (0 to 5 seconds) The rocket moves upward with a constant acceleration of $20\ \mathrm{m/s^2}$.
    2. Phase 2: Upward motion under gravity (after 5 seconds until maximum height) After 5 seconds, the fuel runs out, so the rocket stops accelerating due to its engine. From this point, only gravity acts on it, causing it to decelerate as it continues to move upward until its velocity becomes zero at the maximum height.
    3. Phase 3: Free fall (from maximum height to the ground) Once it reaches its maximum height, the rocket starts falling back to Earth under the influence of gravity. We need to find its speed just before it hits the ground.

    We will use the standard kinematic equations for constant acceleration:

    $v = u + at$

    $s = ut + \frac{1}{2}at^2$

    $v^2 = u^2 + 2as$

    where $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration, $t$ is time, and $s$ is displacement. We will take the upward direction as positive for the first two phases and then adjust for the third phase.

    Let's calculate the motion step-by-step:

    1. Phase 1: Rocket accelerating upwards (from $t=0$ to $t=5\ \mathrm{s}$)

       The rocket starts from rest, so its initial velocity is $u_1 = 0\ \mathrm{m/s}$. It accelerates upward at $a_1 = 20\ \mathrm{m/s^2}$ for $t_1 = 5\ \mathrm{s}$.

       We first find the velocity of the rocket at the end of this phase:

       $v_1 = u_1 + a_1 t_1$

       $v_1 = 0 + (20\ \mathrm{m/s^2})(5\ \mathrm{s})$

       $v_1 = 100\ \mathrm{m/s}$ (upward)

       Next, we find the height reached during this phase:

       $h_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2$

       $h_1 = (0)(5\ \mathrm{s}) + \frac{1}{2} (20\ \mathrm{m/s^2})(5\ \mathrm{s})^2$

       $h_1 = 0 + \frac{1}{2} (20)(25)\ \mathrm{m}$

       $h_1 = 10 \times 25\ \mathrm{m}$

       $h_1 = 250\ \mathrm{m}$

    2. Phase 2: Rocket moving upwards under gravity (after $t=5\ \mathrm{s}$ until maximum height)

       At $t=5\ \mathrm{s}$, the rocket's engine stops, but it has an upward velocity of $100\ \mathrm{m/s}$. From this point, the only acceleration acting on it is due to gravity, which is $g = 10\ \mathrm{m/s^2}$ downward. So, the acceleration for this phase is $a_2 = -10\ \mathrm{m/s^2}$ (taking upward as positive). The rocket will continue to move upward until its final velocity becomes $v_2 = 0\ \mathrm{m/s}$ at the maximum height.

       Using the equation $v^2 = u^2 + 2as$: …

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