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

Step 1 — Find the mass from the first case.

Acceleration in the first case:

Using $F = ma$:

Step 2 — Acceleration in the second case.

Step 3 — Required force.

The mass is the same body, so

We are given $F = \dfrac{A}{B + x^{1.5}}$, where $F$ is force, $x$ is distance, and $A$ and $B$ are constants. The dimensional formula of $AB$ is $[M^a L^b T^c]$, and we need $a+b+c$.

Concept and intuition:

The fundamental rule in dimensional analysis is that you can only add or subtract quantities that have the same dimensions. Here, $B$ and $x^{1.5}$ are added, so they must share the same dimensional formula. That lets us find the dimensions of $B$ directly from $x$. Then, since the whole fraction equals force, we can solve for the dimensions of $A$. Multiplying the dimensions of $A$ and $B$ gives the dimensions of $AB$, and we simply add the exponents.

Step-by-step reasoning:

1. Identify dimensions of known quantities.

   Force $F$ has dimensions $[M L T^{-2}]$.

   Distance $x$ has dimensions $[L]$.

2. Use the addition rule for $B + x^{1.5}$.

   Since $B$ and $x^{1.5}$ are added, they must have the same dimensions.

   $x^{1.5}$ has dimensions $[L]^{1.5} = [L^{3/2}]$.

   Therefore, $B$ also has dimensions $[L^{3/2}]$.

3. Find dimensions of $A$ from the force equation.

   The equation $F = \dfrac{A}{B + x^{1.5}}$ implies that the denominator has the same dimensions as $A$ divided by $F$.

   So, $[A] = [F] \times [B + x^{1.5}]$.

   Since $B + x^{1.5}$ has dimensions $[L^{3/2}]$, we have:

   $[A] = [M L T^{-2}] \times [L^{3/2}] = [M L^{1 + 3/2} T^{-2}] = [M L^{5/2} T^{-2}]$.

4. Determine dimensions of $AB$.

   $[AB] = [A] \times [B] = [M L^{5/2} T^{-2}] \times [L^{3/2}] = [M L^{(5/2 + 3/2)} T^{-2}] = [M L^{4} T^{-2}]$.

   So $a = 1$, $b = 4$, $c = -2$.

5. Compute $a + b + c$.

   $a + b + c = 1 + 4 + (-2) = 3$.

The key insight is that "constant velocity" means no net force along the incline. For a block sliding down a rough incline at constant speed, the frictional force must exactly oppose and equal the component of gravity pulling it down. That tells us the magnitude of friction — and when we pull the block up, friction still acts opposite to motion (now down the plane), so the external force has to overcome both gravity and friction.

Let’s work through it step by step.

1. Set up the forces for downward motion. The block of mass $m = 3\,\text{kg}$ moves down with constant velocity, so the net force along the incline is zero. Let $\theta$ be the angle of the incline. The component of weight down the plane is $mg\sin\theta$. Friction $f$ acts up the plane (opposing motion). Hence:

2. Work done by gravity when the block descends a height $h$.

   The block moves down along the incline through a vertical height $h = 50\,\text{cm} = 0.5\,\text{m}$. The distance along the incline $s$ is related by $h = s\sin\theta$, so $s = \frac{h}{\sin\theta}$.

   Gravity does work $mgh$ (positive because displacement is downward). But we don’t need that directly — we need the work to pull it up.

3. Work required to pull the block up the same incline through the same height.

   When pulling upward at constant speed (again constant velocity, so no acceleration), the external force $F$ must balance both the component of gravity down the plane ($mg\sin\theta$) and friction ($f$), which now also acts down the plane (opposing upward motion). So:

4. Work done by the external force. The displacement along the incline is $s = \frac{h}{\sin\theta}$. Therefore:

Notice the $\sin\theta$ cancels — the result is independent of the incline angle.

5. Plug in the numbers. $m = 3\,\text{kg}$, $g = 10\,\text{m/s}^2$, $h = 0.5\,\text{m}$:

Concept. Newton's second law gives the (uniform) acceleration produced by a constant force; the kinematic relation $v^2 = u^2 + 2as$ then connects the change in speed to the displacement, with no need to know the time.

Step 1 — acceleration.

Step 2 — displacement from the work–energy style kinematic equation.

Check (work–energy theorem). $\Delta KE = \tfrac{1}{2}(6)(36-16) = 60$ J; $W = Fs = 12 \times 5 = 60$ J. Consistent.

ANSWER: B

The key concept is the Lorentz force law for a charge moving in a magnetic field:

The force is perpendicular to both the velocity and the magnetic field. That means the direction of $\vec{F}$ is given by the cross product, and its magnitude depends on the sine of the angle between $\vec{v}$ and $\vec{B}$.

Here, the velocity is purely along the x-axis, while the magnetic field has components in all three directions. The cross product will tell us exactly which plane the force lies in.

1. Write down the vectors

   Velocity: $\vec{v} = 2\hat{i} \ \mathrm{m/s}$

   Magnetic field: $\vec{B} = 2\hat{i} + 2\hat{j} + 3\hat{k} \ \mathrm{T}$

2. Compute the cross product $\vec{v} \times \vec{B}$

Expand:

3. Interpret the result

   The cross product has no $\hat{i}$ component — it lies entirely in the y–z plane. Since $\vec{F} = q(\vec{v} \times \vec{B})$, the force vector is parallel to this cross product (scaled by $q$). So the force is also in the y–z plane.

4. Check the options

   • (A) In y–z plane → True
   • (B) Along –y axis → No, it has both –y and +z components
   • (C) Along +z axis → No, it has a –y component too
   • (D) Along –z axis → No, it has +z component

The key idea here is that static friction is a self-adjusting force. It does not always act at its maximum value — it simply matches whatever force tries to slide the object, up to a limit. So you must first check whether the block would slide if friction were absent, then see if the available maximum static friction can prevent that sliding.

Let’s work through it.

1. Find the component of weight pulling the block down the incline. The weight is $mg = 2 \times 10 = 20\,\text{N}$. The component parallel to the incline is

This is the force trying to slide the block downhill.

2. Find the maximum static friction available. The normal reaction on the incline is

The maximum static friction is

Numerically, $7\sqrt3 \approx 12.12\,\text{N}$.

3. Compare the downhill force with the maximum friction.

   The downhill force is $10\,\text{N}$, which is less than $7\sqrt3 \approx 12.12\,\text{N}$.

   So the block does not move. Static friction does not need to act at its maximum — it only needs to supply enough force to cancel the downhill pull.

4. Therefore, the actual frictional force equals the downhill component.

It acts up the incline, opposing the tendency to slide down.

The key idea is that light carries momentum, and when it hits a surface, the change in momentum per second gives the force. For a perfectly absorbing surface, the force is $P/c$; for a perfectly reflecting surface, it's $2P/c$. Here, the surface does both — 70% absorption and 30% reflection — so we need to combine the contributions.

Think of it this way: each photon that gets absorbed transfers its full momentum to the surface. Each photon that gets reflected reverses its momentum, so the surface gets twice the momentum kick. The total force is just the sum of these two effects, weighted by the fractions of power absorbed and reflected.

1. Find the incident power and momentum flux. The incident beam carries power $P = 10$ W. The momentum per second (momentum flux) carried by the light is $P/c$, since each photon of energy $E$ has momentum $E/c$. Here $c = 3 \times 10^8$ m/s. So the incident momentum per second is

2. Contribution from the absorbed part (70%). For absorption, the surface stops the light completely. The change in momentum per second for the absorbed portion is equal to the momentum it carried, since final momentum is zero. Force from absorption = (fraction absorbed) × (incident momentum per second)

3. Contribution from the reflected part (30%). For reflection, the light bounces back with equal speed but opposite direction. The change in momentum is twice the incident momentum (from $+p$ to $-p$, so change = $2p$). Force from reflection = (fraction reflected) × $2 \times$ (incident momentum per second)

4. Add the forces. Both forces act in the same direction (pushing the surface forward), so they add:

Concept and intuition

We have two separate situations. In the first, a known force on block A alone gives us its acceleration — so we can find the mass of A. In the second, the same block A is pushed against block B, and a different force is applied to the combination. Since friction is neglected, the two blocks move together as one system. The force that block B experiences is simply the net force needed to accelerate block B at the system’s common acceleration. This is a classic “two-block pushed together” problem: find the system acceleration, then isolate one block.

1. Find the mass of block A From the first scenario:

So

2. Treat A and B as a single system under the new force The total mass is

The applied force is $F' = 20~\mathrm{N}$. The common acceleration of the system (no friction) is

3. Find the force on block B Block B is accelerated only by the contact force from block A (call it $F_{AB}$). Applying Newton’s second law to block B alone:

This is the force that block A exerts on block B — which is the force “on block B” asked in the problem.

Given: firing rate $n = 5\ \text{s}^{-1}$, bullet mass $m = 5\ \text{g} = 0.005\ \text{kg}$, holding force $F = 12.5\ \text{N}$.

Speed. The average force needed to hold the gun equals the momentum delivered to the bullets each second:

Power. Kinetic energy of one bullet:

Energy is imparted to bullets at the firing rate, so the power delivered is

Thus $S = 500\ \text{m/s}$ and $P = 3.125\ \text{kW}$.

The work done by a variable force equals the area under the $F$-vs-$x$ curve: when the force is not constant you integrate $F\,dx$, which geometrically is the area between the curve and the $x$-axis. The graph here is piecewise linear, so we split it into simple shapes.

1. From $x=0$ to $2$ m the force rises linearly from $0$ to $5$ N — a triangle:

2. From $x=2$ to $5$ m the force is constant at $5$ N — a rectangle:

3. Total work:

Concept and intuition:

When a constant force acts over a time interval, it changes the object’s momentum. The impulse–momentum theorem says:

Here the object reverses direction, so the change in velocity is large — and the force must be opposite to the original motion. We don’t need to know acceleration or distance; just the start and end velocities, the mass, and the time.

Step-by-step reasoning:

1. Choose a sign convention.

   Let the initial direction of motion be positive.

   So initial velocity: $v_i = +15\ \mathrm{m/s}$.

   The final velocity is $5\ \mathrm{m/s}$ in the opposite direction, so $v_f = -5\ \mathrm{m/s}$.

2. Write the impulse–momentum equation.

   For a constant force $F$ acting for time $\Delta t$:

3. Substitute the known values. $m = 15\ \mathrm{kg}$, $\Delta t = 5\ \mathrm{s}$, $v_i = +15$, $v_f = -5$:

4. Simplify the velocity change.

So:

5. Solve for $F$.

The negative sign means the force acts opposite to the initial direction.