Q.A block of mass $m = 1\ \text{kg}$, moving on a horizontal surface with speed $v_i = 2\ \text{m s}^{-1}$ enters a rough patch ranging from $x = 0.10\ \text{m}$ to $x = 2.01\ \text{m}$. The retarding force $F_r$ on the block in this range is inversely proportional to $x$ over this range,

[!FORMULA] $F_r = -\frac{k}{x} \quad \text{for } 0.1 < x < 2.01\ \text{m}$

[!FORMULA] $= 0 \quad \text{for } x < 0.1\ \text{m and } x > 2.01\ \text{m}$

where $k = 0.5\ \text{J}$. What is the final kinetic energy and speed $v_f$ of the block as it crosses this patch?

Step 1 — The concept: potential difference is work per unit charge

The potential difference between two points is defined as the work done by an external agent in carrying a unit positive charge from one to the other, slowly (without changing its kinetic energy):

Rearranged, this gives the working formula:

Equivalently, $W$ equals the change in electrostatic potential energy, $W = \Delta U = U_A - U_B = q V_A - q V_B$. Since electrostatic force is conservative, this depends only on the endpoints — the path taken is irrelevant.

Step 2 — Identify the initial and final points (order matters!)

The charge is carried from B to A. So:

• Initial point = B, with $V_B = +5$ V
• Final point = A, with $V_A = -3$ V

The potential difference we need is $V_{\text{final}} - V_{\text{initial}} = V_A - V_B$ — not the other way round. Reversing this is the single most common error here and simply flips the sign.

Step 3 — Convert the charge to SI units

(The prefix milli = $10^{-3}$. Forgetting this conversion gives $-40$ J — option (D), a deliberate trap. Mistaking mC for μC would give $-4\times10^{-5}$ J.)

Step 4 — Substitute

Step 5 — Why is the sign negative? (Check the physics)

The negative sign is not an accident — it carries real meaning. …

Concept & Intuition

When a charged particle is released from rest in a uniform electric field, the field exerts a constant force $F = qE$ on it. This force does work as the particle moves, and that work is entirely converted into kinetic energy (since no other forces act). The work done by a constant force over a displacement $d$ is simply $Fd$. So the kinetic energy after moving distance $d$ is $K = qEd$. No need to involve mass or acceleration — the energy approach is direct.

Step-by-step solution

1. Identify the given quantities

   • Mass $m = 3 \, \text{g} = 3 \times 10^{-3} \, \text{kg}$ (not needed for energy, but given)
   • Charge $q = 60 \, \mu\text{C} = 60 \times 10^{-6} \, \text{C} = 6 \times 10^{-5} \, \text{C}$
   • Electric field $E = 10^5 \, \text{N/C}$
   • Distance moved $d = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}$

2. Work done by the electric field

   The force on the particle is $F = qE$. Since the field is uniform, the force is constant. The work done over displacement $d$ is:

Substitute the values:

3. Simplify step by step First, $6 \times 10^{-5} \times 10^5 = 6 \times 10^{0} = 6$. Then multiply by $2 \times 10^{-2}$: $$$$

Concept. As the ball drops onto the spring and compresses it, gravity does positive work over the whole fall, and part of that energy is stored as elastic potential energy in the spring. The net work done in the process is the work delivered by gravity minus the energy locked up in the spring.

Step 1 — Total vertical drop.

The ball starts $2\,$m above the free (upper) end of the spring and the spring is compressed by $50\,\text{cm}=0.5\,$m, so the ball descends

Step 2 — Work done by gravity.

With $m=0.5\,$kg and $g=10\,\text{m s}^{-2}$,

$W_{\text{grav}} = mgh = 0.5\times 10 \times 2.5 = 12.5\ \text{J}.$ …

The key idea is that the brakes apply a constant resisting force (the total resistance to motion) over the stopping distance. The work done by that force equals the change in kinetic energy. Because the weight $W$ is given, we can express the mass as $m = W/g$ and then solve for the force in terms of $W$.

1. Identify the known quantities

   Initial speed: $u = 7\ \mathrm{m/s}$

   Stopping distance: $s = 10\ \mathrm{m}$

   Final speed: $v = 0$

   Weight of scooter: $W$ (in newtons)

   Acceleration due to gravity: $g = 9.8\ \mathrm{m/s^2}$ (we will keep it symbolic initially).

2. Relate force to acceleration using kinematics

   From $v^2 = u^2 + 2as$, we get

The negative sign means deceleration. The magnitude of the acceleration is $a = 2.45\ \mathrm{m/s^2}$.

3. Apply Newton’s second law The total resisting force $F$ (assumed constant) is

where $m$ is the mass. Since weight $W = mg$, we have $m = W/g$. Thus

4. Substitute numbers Using $g = 9.8\ \mathrm{m/s^2}$ and $a = 2.45\ \mathrm{m/s^2}$: $$$$