A body of mass m = 10^-2 kg is moving in a medium and experiences a frictional force F = -kv^2. Its

Q: Why do we use $dv/dt$ for acceleration instead of $v dv/dx$?

Because the problem gives us information about the time elapsed ($t=10$ s), not the distance covered. If distance was given, $v dv/dx$ would be the right choice!

Q: Why is there a negative sign in the force equation $F = -kv^2$?

The negative sign indicates that the force is resistive—it acts in the direction opposite to the velocity, actively slowing the body down.

Q: Can we use the standard kinematic equation $v = u + at$?

Absolutely not! That equation is strictly for constant acceleration. Here, acceleration depends on velocity ($a \propto v^2$), so it changes continuously as the body slows down.

Q: What happens to the kinetic energy that is lost?

The lost kinetic energy is dissipated into the surrounding medium as heat and sound due to the work done against the frictional drag force.

Q: Why did we take the positive square root when finding the final velocity?

The body is simply slowing down due to drag; it hasn't reversed its direction of motion. Therefore, the velocity remains positive.

Question

A body of mass $m = 10^{-2} 	ext{ kg}$ is moving in a medium and experiences a frictional force $F = -kv^2$. Its initial speed is $v_0 = 10 	ext{ ms}^{-1}$. If, after 10 s, its energy is $\frac{1}{8}mv_0^2$, the value of $k$ will be

Options:

(A) $10^{-3} 	ext{ kgs}^{-1}$

(B) $10^{-4} 	ext{ kgm}^{-1}$

(C) $10^{-1} 	ext{ kgm}^{-1}	ext{s}^{-1}$

(D) $10^{-3} 	ext{ kgm}^{-1}$

Sigma Percentile · Accepted Answer

### The Setup: Battling the Drag

Imagine a block moving through a thick, viscous fluid. Unlike standard friction which is often constant, fluid drag is dynamic—it grows stronger as you move faster. In this problem, the medium exerts a resistive force that is proportional to the square of the velocity, given by $F = -kv^2$. The negative sign is nature's way of telling us that this force is actively opposing the motion, trying to bring the block to a halt.

### The Calculus of Motion

To understand how the block slows down over time, we must call upon Newton's second law, $F = ma$. Since acceleration is the rate of change of velocity, we can write:

$$m \frac{dv}{dt} = -kv^2$$

This is a classic separable differential equation. To solve it, we group all the velocity terms on one side and the time terms on the other:

$$\frac{dv}{v^2} = -\frac{k}{m} dt$$

Now, we integrate both sides. The clock starts at $t = 0$ with an initial velocity $v_0 = 10 	ext{ m/s}$, and we want to find the state at $t = 10 	ext{ s}$ where the velocity is $v$:

$$\int_{v_0}^{v} v^{-2} dv = -\frac{k}{m} \int_{0}^{10} dt$$

Evaluating the integral of $v^{-2}$ yields $-1/v$. Plugging in the limits gives us a beautiful algebraic relationship:

$$\frac{1}{v} - \frac{1}{v_0} = \frac{10k}{m}$$

### The Energy Clue

We have an equation, but we are missing the final velocity $v$. Fortunately, the problem provides a crucial clue: after 10 seconds, the kinetic energy of the block drops to exactly $\frac{1}{8}$ of its initial value.

Let's translate this physical condition into math:

$$K_f = \frac{1}{8} K_i$$

$$\frac{1}{2} m v^2 = \frac{1}{8} \left( \frac{1}{2} m v_0^2 ight)$$

The mass $m$ and the $\frac{1}{2}$ factor cancel out perfectly, leaving us with:

$$v^2 = \frac{v_0^2}{4}$$

Taking the square root, we find that the final velocity is exactly half of the initial velocity. Since $v_0 = 10 	ext{ m/s}$, our final velocity is $v = 5 	ext{ m/s}$.

### The Final Calculation

We now have all the pieces of the puzzle. Let's substitute $v = 5 	ext{ m/s}$, $v_0 = 10 	ext{ m/s}$, and $m = 10^{-2} 	ext{ kg}$ back into our integrated equation:

$$\frac{1}{5} - \frac{1}{10} = \frac{10k}{10^{-2}}$$

Simplifying the left side, $\frac{1}{5} - \frac{1}{10}$ is simply $\frac{1}{10}$. On the right side, dividing by $10^{-2}$ is equivalent to multiplying by $100$, so $10 / 10^{-2} = 1000$.

$$\frac{1}{10} = 1000k$$

Solving for $k$, we get:

$$k = 10^{-4} 	ext{ kg m}^{-1}$$

And there we have it! By seamlessly blending Newton's laws, calculus, and the work-energy theorem, we've decoded the exact nature of the drag force.

Animated Solution for Physics - Work, Energy, and Power: A body of mass $m = 1 0^{- 2} kg$ is moving in a medium and experiences a frictional force $F = - k v^{2}$ . Its initial speed is $v_{0} = 10 ms^{- 1}$ . If, after 10 s, its energy is $\frac{1}{8} m v_{0}^{2}$ , the value of $k$ will be

Select Answer:

Visualized Solution

Visualizing the Drag Force

Applying Newton's Second Law

Separating Variables

Setting Up the Integral

Evaluating the Integral

Using the Energy Condition

Calculating Final Velocity

Substituting Known Values

Final Calculation

The Way Forward

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Setup

The Calculus of Motion

The Energy Clue

The Final Calculation

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