A marble block of mass 2 kg lying on ice when given a velocity of 6 ms^-1 is stopped by friction in

Q: Why didn't we use the mass of 2 kg in our calculation?

The deceleration caused by kinetic friction on a horizontal surface is $a = -\mu g$. Since the frictional force $f = \mu m g$ and $f = ma$, the mass $m$ cancels out on both sides of the equation. The stopping time is completely independent of mass!

Q: Why is the acceleration negative?

Friction always opposes the relative motion of the block. Since we took the initial velocity as positive (moving forward), the acceleration (which slows the block down) must be in the opposite direction, hence it is negative.

Q: What if the block was moving on an inclined plane?

On an inclined plane, the normal force changes to $mg \cos\theta$, and gravity also pulls the block down the incline with a force of $mg \sin\theta$. The deceleration would be $g(\sin\theta + \mu \cos\theta)$ if moving up, or $g(\mu \cos\theta - \sin\theta)$ if moving down.

Q: Does the coefficient of friction have units?

No, the coefficient of friction $\mu$ is a dimensionless quantity. It is defined as the ratio of the frictional force to the normal force ($f/N$), both of which are measured in Newtons. The units cancel out.

Q: Why did we take $g = 10 \text{ m/s}^2$?

In many competitive exams like JEE and NEET, $g$ is approximated as $10 \text{ m/s}^2$ to simplify calculations and avoid messy decimals, unless $9.8 \text{ m/s}^2$ is explicitly specified in the problem.

Question

A marble block of mass $2	ext{ kg}$ lying on ice when given a velocity of $6	ext{ ms}^{-1}$ is stopped by friction in $10	ext{ s}$. Then, the coefficient of friction is

Options:

(A) $0.02$

(B) $0.03$

(C) $0.06$

(D) $0.01$

Sigma Percentile · Accepted Answer

### The Setup: A Block on Ice

Imagine a marble block sliding across a vast, smooth sheet of ice. It begins its journey with a brisk initial velocity of $u = 6 	ext{ m/s}$. However, the ice isn't perfectly frictionless. A subtle kinetic frictional force acts in the opposite direction of motion, gradually sapping the block's kinetic energy until it comes to a complete halt ($v = 0 	ext{ m/s}$) in exactly $t = 10 	ext{ s}$.

Our goal is to uncover the hidden property of the surface: the coefficient of kinetic friction, $\mu$.

### The Master Equation: Kinematics Meets Dynamics

To solve this, we must bridge the gap between the block's motion (kinematics) and the forces acting upon it (dynamics). The block is undergoing uniform deceleration. The perfect tool for this is the first equation of motion:

$$v = u + at$$

But what is the acceleration $a$? From Newton's Second Law, the net force acting horizontally is solely the kinetic friction $f_k$. We know that $f_k = \mu N$, and on a flat horizontal surface, the normal force $N$ perfectly balances gravity, so $N = mg$.

Therefore, the frictional force is $f_k = \mu mg$. Since this force opposes motion, it creates a deceleration (negative acceleration):

$$ma = -\mu mg \implies a = -\mu g$$

### The Illusion of Mass

Now, let's substitute our known values into the kinematic equation:

$$0 = 6 + (-\mu g)(10)$$

Notice something fascinating here? The mass of the block, given as $2 	ext{ kg}$, is completely absent from our equation! This is a classic trap designed to test your conceptual clarity. The deceleration caused by friction on a horizontal surface is independent of the object's mass. A massive boulder and a tiny pebble, if given the same initial speed on the same surface, will slide for the exact same duration before stopping.

### The Final Calculation

Let's rearrange our equation to isolate the unknown variable, $\mu$:

$$10 \mu g = 6$$

$$\mu = \frac{6}{10g}$$

Taking the standard approximation for the acceleration due to gravity, $g = 10 	ext{ m/s}^2$, we get:

$$\mu = \frac{6}{10 	imes 10} = \frac{6}{100}$$

$$\mu = 0.06$$

And there we have it! The coefficient of friction between the marble block and the ice is a mere **$0.06$**, confirming that the ice is indeed quite slippery, but not entirely frictionless.

Animated Solution for Physics - Laws of Motion: A marble block of mass $2 kg$ lying on ice when given a velocity of $6 ms^{- 1}$ is stopped by friction in $10 s$ . Then, the coefficient of friction is

Select Answer:

Visualized Solution

\text{Visualizing the Motion}

\text{Kinematics \& Dynamics}

\text{Substituting Values}

\text{Solving for } \mu

\text{Final Calculation}

\text{Mass Independence}

The Sigma Insight: Static and Kinetic Friction

The Setup

The Master Equation

The Illusion of Mass

The Final Calculation

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