Q.Give the magnitude and direction of the net force acting on a stone of mass $0.1\ \text{kg}$,

Showing the 12 most recent of 36 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Two blocks of masses m and M are connected by an inextensible light string [figure: block m rests on the ground on the left; a string runs from the top of block m up at an angle θ to the horizontal to the top of block M on the right]. A constant horizontal force f acts on the block of mass M, then tension in the string is (Neglect friction) (A) $\dfrac{mf}{(M+m)\cos\theta}$ (B) $\dfrac{Mf}{m\cos\theta}$ (C) $\dfrac{Mf}{(M+m)\cos\theta}$ (D) $\dfrac{mf}{M\cos\theta}$

   ›Reveal solutionSolution

   This tests constrained motion with an inclined connector: since the blocks stay on the ground, only the horizontal component of tension does any accelerating work on $m$. Treat the system together first, then isolate a block. Answer: (A).

   Concept and Intuition

   Both blocks slide on the same horizontal floor, so both must have the same horizontal acceleration $a$ (the string is inextensible and stays taut). The string leaves the top of $m$ at angle $\theta$ to reach the top of $M$, so tension acts along this slanted direction at each end. But block $m$ cannot move vertically — the floor's normal force silently absorbs the vertical component $T\sin\theta$ of the pull on $m$. Only the horizontal component $T\cos\theta$ actually accelerates $m$ along the floor. This is the key simplification: the incline of the string matters only for splitting $T$ into components, not for the kinematics, because both blocks still move purely horizontally together.

   Step-by-Step Solution

   1. Whole system: the only external horizontal force is $f$ (friction is neglected, and the vertical forces — weights, normal reactions, vertical component of $T$ on each block — cancel out with the floor). So

   $f=(M+m)a \quad\Rightarrow\quad a=\frac{f}{M+m}.$

   2. Isolate block $m$: the only horizontal force on $m$ is the horizontal component of the tension, $T\cos\theta$ (there is no friction, and the vertical component $T\sin\theta$ is balanced by the floor's normal reaction, which adjusts to keep $m$ on the ground). Newton's second law horizontally:

   $T\cos\theta = ma.$

   3. Substitute $a$:

   $T=\frac{ma}{\cos\theta}=\frac{m}{\cos\theta}\cdot\frac{f}{M+m}=\frac{mf}{(M+m)\cos\theta}.$

   Common Mistakes

   • Writing the equation for $M$ using $T$ instead of $T\cos\theta$ (forgetting the string is inclined for $M$ too — but since we solved via the system + block $m$, this never needs to be done separately; if you check it as a consistency test, $f-T\cos\theta=Ma$ must also hold).
   • Including $T\sin\theta$ in $m$'s horizontal equation — it's a vertical component and plays no role in horizontal acceleration.
   • Forgetting that both blocks must share the same acceleration only because both remain on the (same) floor — this is what lets you add the equations for the whole system cleanly.

   ✓Final answer

   The correct option is (A) — $\dfrac{mf}{(M+m)\cos\theta}$.

   ANSWER: A

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A block of mass 2 kg placed on a rough horizontal surface is pulled with a force of 30 N which makes an angle of $\sin^{-1}(0.6)$ with the horizontal. If the coefficient of kinetic friction between the block and the surface is 0.4, then the acceleration of the block is (Acceleration due to gravity = $10\ ms^{-2}$) (A) $8.8\ ms^{-2}$ (B) $11.6\ ms^{-2}$ (C) $5.8\ ms^{-2}$ (D) $10.6\ ms^{-2}$

   ›Reveal solutionSolution

   Since the pull has an upward component, it reduces the normal force (and hence friction) below $mg$. Careful force resolution gives $a=11.6\ \text{ms}^{-2}$.

   Concept and Intuition

   When a force is applied at an angle above the horizontal to drag a block, its vertical component partially lifts the block, reducing the normal reaction from the surface (compared to a purely horizontal pull). Since kinetic friction is $\mu N$, a smaller $N$ means less friction opposing the motion — this must be accounted for before computing net horizontal force.

   Step-by-Step Solution

   1. $\theta=\sin^{-1}(0.6)\Rightarrow \sin\theta=0.6,\ \cos\theta=0.8$ (3-4-5 triangle).
   2. Horizontal component of applied force: $F_x = 30\cos\theta = 30(0.8) = 24$ N.
   3. Vertical (upward) component: $F_y = 30\sin\theta = 30(0.6)=18$ N.
   4. Vertical equilibrium (block stays on surface): $N + F_y = mg \Rightarrow N = mg - F_y = (2)(10) - 18 = 20-18=2$ N.
   5. Kinetic friction: $f = \mu N = 0.4(2) = 0.8$ N.
   6. Newton's second law horizontally: $F_x - f = ma \Rightarrow 24 - 0.8 = 2a \Rightarrow a = \dfrac{23.2}{2}=11.6\ \text{ms}^{-2}$.

   Common Mistakes

   • Using $N=mg$ (ignoring the vertical component of the applied force) — this is the classic trap that this question is built to catch.
   • Adding rather than subtracting $F_y$ from $mg$ when the pull is above the horizontal (lifting, not pressing down).

   ✓Final answer

   The correct option is (B) — $11.6\ ms^{-2}$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A body of mass 5 kg at rest is acted upon by two forces 30 N and F. If the angle between the forces is $60^0$ and the distance travelled by the body in a time of 2s under the action of these forces is 28m, then F = (A) 60N (B) 40N (C) 50N (D) 30N

   ›Reveal solutionSolution

   The kinematics gives the net (resultant) force from the distance-time data; the law of vector addition (parallelogram law / law of cosines) then lets us solve for the unknown force $F$. The answer is 50 N.

   Concept and Intuition

   A body starting from rest under a constant net force undergoes uniform acceleration, so $s=\tfrac12at^2$ gives us the acceleration directly from the motion data — no need to know the individual forces to find the resultant force ($R=ma$). Once we know $R$, the two individual forces (30 N and $F$) at a known angle combine via the standard $R^2=F_1^2+F_2^2+2F_1F_2\cos\theta$ relation (same as the law of cosines for the resultant of two vectors).

   Step-by-Step Solution

   1. Body starts from rest ($u=0$); distance $s=28$ m in $t=2$ s. Using $s=ut+\tfrac12at^2$: $28 = 0+\tfrac12 a(4) = 2a \Rightarrow a=14\ \text{ms}^{-2}$.
   2. Resultant force: $R=ma=5(14)=70$ N.
   3. Law of cosines for two forces at angle $60^\circ$: $R^2 = 30^2+F^2+2(30)(F)\cos60^\circ = 900+F^2+30F$ (since $\cos60^\circ=0.5$).
   4. Substitute $R=70$: $4900 = 900+F^2+30F \Rightarrow F^2+30F-4000=0$.
   5. Solve the quadratic: $F=\dfrac{-30\pm\sqrt{900+16000}}{2}=\dfrac{-30\pm\sqrt{16900}}{2}=\dfrac{-30\pm130}{2}$. Taking the positive root: $F=\dfrac{100}{2}=50$ N.

   Common Mistakes

   • Forgetting the initial-rest condition and using a wrong kinematic equation.
   • Sign/arithmetic slip solving the quadratic, or taking the negative (unphysical) root.

   ✓Final answer

   The correct option is (C) — 50N.

   ANSWER: C

4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.3 kg block on rough incline $37^\circ$ is connected to a hanging mass of 4 kg. If the coefficient of friction between the 3 kg block and the rough incline $\mu = 0.25$, then acceleration of the system is [FIGURE: a 3 kg block resting on a frictional incline of angle $37^\circ$, connected by a string over a pulley at the top of the incline to a hanging 4 kg mass] $\left[g=10\ ms^{-2},\ \sin37^\circ=0.6,\ \cos37^\circ=0.8\right]$ (A) 2.28 $ms^{-2}$ (B) 1.08 $ms^{-2}$ (C) 3.2 $ms^{-2}$ (D) Zero

   ›Reveal solutionSolution

   A connected system (block on rough incline + hanging mass over a pulley) is solved with Newton's second law for the system as a whole; the hanging mass wins the tug-of-war and the acceleration comes out to $\approx 2.28\ ms^{-2}$.

   Concept and Intuition

   The string constrains both masses to have the same magnitude of acceleration. To find the direction of motion, we first compare the 'driving' force (weight of the hanging mass) against the 'resisting' forces on the incline (gravity component + friction). Since $40\ N > 18\ N + 6\ N$, the hanging mass falls and the block is dragged up the incline, so friction acts down the incline (opposing relative motion).

   Step-by-Step Solution

   1. Forces on hanging 4 kg mass (taking its falling direction as positive): weight $= 4\times10 = 40\ N$ downward, tension $T$ upward.
   2. Forces on the 3 kg block along the incline (block accelerates up the incline): gravity component $= 3\times10\times\sin37° = 3\times10\times0.6 = 18\ N$ down the incline; friction $f = \mu N = \mu\, mg\cos37° = 0.25\times3\times10\times0.8 = 6\ N$, acting down the incline (opposing the block's upward motion); tension $T$ up the incline.
   3. Newton's second law for the hanging mass: $40 - T = 4a$.
   4. Newton's second law for the block: $T - 18 - 6 = 3a$.
   5. Add the two equations to eliminate $T$: $40 - 18 - 6 = 7a \Rightarrow 16 = 7a \Rightarrow a = 16/7 \approx 2.286\ ms^{-2}$.
   6. This rounds to the given option 2.28 $ms^{-2}$.

   Common Mistakes

   • Forgetting friction acts opposing relative sliding (i.e., down the incline here, not up).
   • Not checking which way the system actually moves before assigning friction's direction.
   • Using $\sin37°=0.6,\cos37°=0.8$ swapped.

   ✓Final answer

   The correct option is (A) — 2.28 $ms^{-2}$.

   ANSWER: A

5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A solid sphere of mass 2 kg is at rest inside a cube as shown in the figure. [FIGURE] (a cube resting on a horizontal surface, with a solid sphere positioned at its bottom-left inside corner; the cube moves to the right along the x-axis with velocity vector $\vec v$ shown by an arrow) Now the cube moves with a velocity, $\vec v = (5t\hat i + 2t\hat j)\ ms^{-1}$. Here $t$ is time in seconds. If the sphere is at rest with respect to cube, the force exerted by the sphere on the cube is (All surfaces are smooth and take $g = 10ms^{-2}$) (A) $\sqrt{29}\ N$ (B) $29\ N$ (C) $26\ N$ (D) $\sqrt{89}\ N$

   ›Reveal solutionSolution

   The sphere accelerates with the cube; applying Newton's second law along both the horizontal (wall-contact) and vertical (floor-contact) directions gives reaction forces of 10 N and 24 N, whose resultant on the cube is $26$ N.

   Concept and Intuition

   Since the sphere is at rest relative to the cube, it must be accelerating with the cube's acceleration in the ground frame. Two contact forces act on the sphere — the normal reaction from the floor (vertical) and from the left wall (horizontal) — and together with gravity they must produce exactly this acceleration. By Newton's third law, the sphere pushes back on the cube's wall and floor with equal and opposite forces; since these two reaction forces are mutually perpendicular (one horizontal, one vertical), their resultant magnitude on the cube is found via Pythagoras.

   Step-by-Step Solution

   1. Cube's velocity: $\vec v=(5t\hat i+2t\hat j)$ m/s, so its acceleration $\vec a=\dfrac{d\vec v}{dt}=(5\hat i+2\hat j)$ m/s² — constant.
   2. The sphere (mass $m=2$ kg) sits in the bottom-left interior corner, touching the floor (vertical normal force $N_1$, upward) and the left wall (horizontal normal force $N_2$, pushing the sphere in $+x$, away from the wall).
   3. Since the sphere shares the cube's acceleration, apply Newton's second law componentwise:
      • Horizontal ($x$): $N_2=ma_x=2\times5=10$ N.
      • Vertical ($y$): $N_1-mg=ma_y \Rightarrow N_1=m(g+a_y)=2\times(10+2)=24$ N.
   4. By Newton's third law, the sphere exerts a force of magnitude $N_2=10$ N on the wall (directed into the wall, i.e. $-x$) and a force of magnitude $N_1=24$ N on the floor (directed into the floor, i.e. $-y$).
   5. These two reaction forces (on two different, perpendicular surfaces of the cube) combine as a net force on the cube of magnitude $\sqrt{N_1^2+N_2^2}=\sqrt{24^2+10^2}=\sqrt{576+100}=\sqrt{676}=26$ N.

   Common Mistakes

   • Forgetting to add the cube's own acceleration to gravity when computing the vertical normal force (using $N_1=mg$ alone).
   • Not realizing that "force exerted by the sphere on the cube" is the vector sum (Pythagorean resultant) of the two separate contact reactions, not just one of them.

   ✓Final answer

   The correct option is (C) — $26\ N$.

   ANSWER: C

6. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A body of mass 5 kg is acted upon by a force $\bar{F} = (-3i + 4j)N$. If its initial velocity at $t = 0$ is, $\bar{u} = (6i - 12j)\mathrm{ms^{-1}}$, the time at which it will just have a velocity along the $y$-axis is (A) Never (B) 10 sec (C) 2 sec (D) 15 sec

   ›Reveal solutionSolution

   The velocity becomes purely along the y-axis when its x-component vanishes; solving $v_x(t)=0$ gives $t=10$ s.

   Concept and Intuition

   "Velocity along the y-axis" means the velocity vector has zero x-component (and a non-zero y-component). Since the force is constant, the acceleration is constant, so each velocity component varies linearly with time — we just need to find when the x-component crosses zero.

   Step-by-Step Solution

   1. Acceleration components: $\vec a = \vec F/m = \dfrac{(-3\hat i+4\hat j)}{5} = (-0.6\hat i + 0.8\hat j)\ \mathrm{ms^{-2}}$.
   2. Velocity as a function of time: $v_x(t) = u_x + a_xt = 6 - 0.6t$; $v_y(t) = u_y + a_yt = -12+0.8t$.
   3. Set $v_x(t) = 0$: $6-0.6t=0 \implies t = \dfrac{6}{0.6}=10\ \text{s}$.
   4. Check $v_y(10) = -12 + 0.8(10) = -4\ \mathrm{ms^{-1}} \ne 0$, confirming the velocity vector is genuinely along the y-axis (not the zero vector) at $t=10$ s.

   Common Mistakes

   • Forgetting to check that $v_y \ne 0$ at that instant (otherwise the velocity would be zero, not "along the y-axis").
   • Sign errors when dividing the force components by the mass.

   ✓Final answer

   The correct option is (B) — 10 sec.

   ANSWER: B

7. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A body of mass 2kg is at rest. When two forces 3N and 4N act on the body in perpendicular directions simultaneously, magnitude and direction of the resultant acceleration are respectively (A) 2 $\mathrm{ms^{-2}}$, $\mathrm{Tan^{-1}}(3/4)$ with 4 N Force (B) 2.5 $\mathrm{ms^{-2}}$, $\mathrm{Tan^{-1}}(3/4)$ with 3 N Force (C) 2.5 $\mathrm{ms^{-2}}$, $\mathrm{Tan^{-1}}(3/4)$ with 4 N Force (D) 2 $\mathrm{ms^{-2}}$, $\mathrm{Tan^{-1}}(4/3)$ with 3 N Force

   ›Reveal solutionSolution

   Two perpendicular forces (3 N, 4 N) combine to a 5 N resultant, giving acceleration $2.5\ \mathrm{ms^{-2}}$ directed at $\tan^{-1}(3/4)$ from the 4 N force.

   Concept and Intuition

   When two forces act perpendicular to each other, they form the two legs of a right triangle, and the resultant is the hypotenuse — a direct application of vector addition (and Newton's second law to get acceleration).

   Step-by-Step Solution

   1. Magnitude of resultant force: $F = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5\ \text{N}$.
   2. Acceleration: $a = F/m = 5/2 = 2.5\ \mathrm{ms^{-2}}$.
   3. Direction: placing the 4 N force along a reference axis, the 3 N force is perpendicular to it. The angle $\phi$ the resultant makes with the 4 N force satisfies

   $\tan\phi = \frac{\text{opposite (3 N)}}{\text{adjacent (4 N)}} = \frac{3}{4} \implies \phi = \tan^{-1}(3/4)$

   (Measuring the angle relative to the 3 N force instead would give $\tan^{-1}(4/3)$, a different value — so the reference force matters.)

   Common Mistakes

   • Swapping which force is used as the reference axis, mixing up $\tan^{-1}(3/4)$ with $\tan^{-1}(4/3)$.
   • Forgetting to divide the resultant force by mass to get acceleration.

   ✓Final answer

   The correct option is (C) — 2.5 $\mathrm{ms^{-2}}$, $\mathrm{Tan^{-1}}(3/4)$ with 4 N Force.

   ANSWER: C

8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A body of mass 5 kg starts from the origin with an initial velocity $(30\hat{i}+40\hat{j})\ \mathrm{m\,s^{-1}}$. If a constant force $-(\hat{i}+5\hat{j})\ \mathrm{N}$ acts on the body, then the time in which the y-component of its velocity becomes zero is (A) 5 s (B) 20 s (C) 40 s (D) 80 s

   ›Reveal solutionSolution

   With constant force $-(\hat i+5\hat j)$ N on a 5 kg body, the y-acceleration is $-1\ \mathrm{m\,s^{-2}}$; starting from $v_y=40\ \mathrm{m\,s^{-1}}$, it takes (C) 40 s to reach zero.

   Concept and Intuition

   Since force, mass, and initial velocity are all given in component form, we can treat the x- and y-motions completely independently (Newton's second law applies component-wise). We only need the y-component of the force and initial velocity to find when $v_y$ becomes zero — the x-motion is irrelevant to this question.

   Step-by-Step Solution

   1. Given: $m=5$ kg, $\vec u=(30\hat i+40\hat j)\ \mathrm{m\,s^{-1}}$, $\vec F=-(\hat i+5\hat j)$ N.
   2. Acceleration: $\vec a=\vec F/m=\dfrac{-(1,5)}{5}=(-0.2,-1)\ \mathrm{m\,s^{-2}}$.
   3. So $a_y=-1\ \mathrm{m\,s^{-2}}$, constant.
   4. y-velocity as a function of time: $v_y(t)=u_y+a_y t=40-t$.
   5. Set $v_y(t)=0$: $40-t=0\ \Rightarrow\ t=40$ s.

   Common Mistakes

   • Using the x-component of force/velocity by mistake instead of the y-component.
   • Forgetting to divide the force by mass to get acceleration before using $v=u+at$.
   • Sign errors on the force direction (the force is $-\hat j$ direction in the y-component, which correctly decelerates the positive $u_y=40$).

   ✓Final answer

   The correct option is (C) — 40 s.

   ANSWER: C

9. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.Two blocks of masses 8 kg and 12 kg kept on smooth horizontal table are connected to the ends of a light string as shown in the figure. If a horizontal force of 500 N is applied to the block of mass 12 kg, then the tension in the string connecting the blocks is [FIGURE] (two blocks, 8 kg and 12 kg, resting on a horizontal table and joined by a string; a horizontal force of 500 N is applied to the 12 kg block, pulling it away from the 8 kg block) (A) 200 N (B) 300 N (C) 500 N (D) 250 N

   ›Reveal solutionSolution

   The two blocks accelerate together under the 500 N force; applying Newton’s second law to the whole system gives the acceleration, and then isolating the 8 kg block yields the tension. The tension is 200 N, so option (A) is correct.

   ---

   Why this approach works

   When two objects are connected by a light, inextensible string on a smooth (frictionless) table, they must move with the same acceleration. The string is “light” — its mass is negligible — so the tension is the same at both ends.

   Newton’s second law, $F_{\text{net}} = m a$, can be applied in two powerful ways:

   1. To the whole system – this gives the common acceleration directly, because the tension is an internal force and cancels out.
   2. To just one block – this lets us solve for the tension, since we already know the acceleration.

   The key insight: the tension is not the applied force; it is only the force needed to accelerate the 8 kg block at the same rate as the whole system.

   ---

   Step‑by‑step solution

   1. Find the acceleration of the whole system

   The total mass being pulled is

   $m_{\text{total}} = 8\ \text{kg} + 12\ \text{kg} = 20\ \text{kg}.$

   The only external horizontal force on the system is the 500 N pull (the string tension is internal, so it does not appear in the system‑level equation).

   Applying Newton’s second law to the whole system:

   $F_{\text{net}} = m_{\text{total}}\, a \quad\Rightarrow\quad 500\ \text{N} = 20\ \text{kg} \times a.$

   Thus

   $a = \frac{500}{20} = 25\ \text{m/s}^2.$

   Tip

   Because the table is smooth, there is no friction to subtract. If friction were present, we would need to subtract the total friction from 500 N first.

   2. Isolate the 8 kg block to find tension

   Now look only at the 8 kg block. The only horizontal force acting on it is the tension $T$ in the string, pulling it to the right.

   Since it accelerates at $a = 25\ \text{m/s}^2$ (same as the whole system), Newton’s second law gives:

   $T = m_1\, a = 8\ \text{kg} \times 25\ \text{m/s}^2 = 200\ \text{N}.$

   Watch out

   A common mistake is to think the tension equals the applied force (500 N). That would only happen if the 8 kg block were fixed or if the string were not accelerating the 12 kg block. Here, part of the 500 N is used to accelerate the 12 kg block itself, so the tension is smaller.

   3. (Optional) Check with the 12 kg block

   For completeness, apply Newton’s second law to the 12 kg block. The forces on it are:

   • 500 N to the right (applied),
   • Tension $T$ to the left (from the string).

   Net force:

   $500 - T = m_2\, a = 12 \times 25 = 300\ \text{N}.$

   So $500 - T = 300$ gives $T = 200\ \text{N}$, confirming our result.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
    Q.A force of 18 N is acting in the direction of motion of a body of mass 3 kg moving with a velocity of 2 m s$^{-1}$. The velocity of the body when it displaces by 5 m is (A) 4 m s$^{-1}$ (B) 6 m s$^{-1}$ (C) 10 m s$^{-1}$ (D) 8 m s$^{-1}$

    ›Reveal solutionSolution

    This tests the work-energy/kinematics relation for a body under constant force over a given displacement. The answer is (D) 8 m s$^{-1}$.

    Concept and Intuition

    A constant force acting in the direction of motion produces constant acceleration (Newton's second law), and once we know that acceleration, the standard kinematic relation connecting velocity and displacement (which doesn't require knowing time) directly gives the final speed after a given displacement.

    Step-by-Step Solution

    1. Acceleration from the applied force: $a = F/m = 18/3 = 6\text{ m/s}^2$.
    2. Initial velocity: $u = 2\text{ m/s}$; displacement: $s = 5\text{ m}$.
    3. Apply $v^2 = u^2 + 2as = (2)^2 + 2(6)(5) = 4 + 60 = 64$.
    4. So $v = \sqrt{64} = 8\text{ m/s}$.

    Common Mistakes

    • Forgetting to square the initial velocity or mis-applying the kinematic equation (e.g., using $v=u+at$ without first finding $t$, which is unnecessary extra work but should still give the same answer if done correctly).

    ✓Final answer

    The correct option is (D) — 8 m s$^{-1}$.

    ANSWER: D

11. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.A force of 10 N acts on a body of mass 2 kg in the direction of motion of the body. If the velocity of the body at a time t = 0 is $13 \mathrm{~m~s^{-1}}$, then the velocity of the body at time t = 3s is (A) $20 \mathrm{~m~s^{-1}}$ (B) $24 \mathrm{~m~s^{-1}}$ (C) $28 \mathrm{~m~s^{-1}}$ (D) $32 \mathrm{~m~s^{-1}}$

    ›Reveal solutionSolution

    A constant force along the direction of motion produces a constant acceleration; applying $v=v_0+at$ with $a=5\,\mathrm{m/s^2}$ gives $28\,\mathrm{m/s}$ at $t=3\,\mathrm{s}$.

    Concept and Intuition

    Since the applied force acts in the same direction as the body's motion, it simply produces a constant acceleration in that same direction (no need to worry about direction reversal or components). This is a direct application of Newton's second law followed by the first equation of motion.

    Step-by-Step Solution

    1. Compute acceleration: $a = \dfrac{F}{m} = \dfrac{10\,\mathrm{N}}{2\,\mathrm{kg}} = 5\ \mathrm{m\,s^{-2}}$.
    2. Apply the kinematic equation: $v = v_0 + at$.
    3. Substitute: $v = 13 + 5\times 3 = 13+15 = 28\ \mathrm{m\,s^{-1}}$.

    Common Mistakes

    • Forgetting that the force is along the direction of motion (so it simply adds to speed, no vector decomposition needed) and making an arithmetic slip in $5\times3$.

    ✓Final answer

    The correct option is (C) — $28\ \mathrm{m\,s^{-1}}$.

    ANSWER: C

12. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.The minimum force required to stop a body of mass 4 kg moving along a straight line with a velocity of 54 kmph in a distance of 9 m is (A) 75 N (B) 100 N (C) 50 N (D) 25 N

    ›Reveal solutionSolution

    Using $v^2=u^2-2as$ to find the deceleration and then $F=ma$ gives the stopping force as 50 N.

    Concept and Intuition

    A force applied opposite to motion decelerates a body uniformly (constant force ⇒ constant deceleration). Kinematics gives the deceleration needed to stop the body over the given distance, and Newton's second law converts that deceleration into the required force.

    Step-by-Step Solution

    1. Convert the initial speed: $u = 54\ \text{kmph} = 54 \times \dfrac{5}{18} = 15\ \mathrm{m\,s^{-1}}$.
    2. Final speed $v = 0$ (body stops), distance $s = 9$ m.
    3. Use $v^2 = u^2 - 2as$:

    $0 = 15^2 - 2a(9) \implies 2a(9) = 225 \implies a = \frac{225}{18} = 12.5\ \mathrm{m\,s^{-2}}$

    4. Apply Newton's second law: $F = ma = 4 \times 12.5 = 50$ N.

    Common Mistakes

    • Forgetting to convert kmph to m/s before using the kinematic equation.
    • Sign errors: since the body decelerates, $a$ in $v^2=u^2-2as$ should be treated as a deceleration (subtracted), which is exactly what was done above.

    ✓Final answer

    The correct option is (C) — 50 N.

    ANSWER: C