Problem 422 - JEE Physics

Question

A ball of mass \(5 \mathrm{~kg}\) is striding on a plane with initial velocity of \(10 \mathrm{~m} / \mathrm{s}\). If co-efficient of friction between surface and ball is \((1 / 2)\), then before stopping it will describe \(\ldots \ldots\) \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(12.5 \mathrm{~m}\) (B) \(5 \mathrm{~m}\) (C) \(7.5 \mathrm{~m}\) (D) \(10 \mathrm{~m}\)

Step-by-Step Solution

Answer

The ball will describe a distance of \(10 \mathrm{~m}\) before stopping.

Step 1: Find the Friction Force

First, we need to calculate the friction force acting on the ball. This force can be determined using the following formula: \[ F_{friction} = \mu \cdot F_{normal} \] Where \(F_{friction}\) is the friction force, \(\mu\) is the coefficient of friction, and \(F_{normal}\) is the normal force. Since the ball is moving on a horizontal surface, the normal force equals the gravitational force acting on it: \[ F_{normal} = m \cdot g \] Now, we can plug in the given values and find the friction force: \[ F_{friction} = \frac{1}{2} \cdot 5 \mathrm{~kg} \cdot 10 \mathrm{~m/s^2} = 25 \mathrm{~N} \]

Step 2: Calculate the Deceleration

Next, we will find the deceleration caused by the friction force acting on the ball. To do this, we'll use Newton's second law of motion: \[ F = m \cdot a \] Where \(F\) is the net force acting on the object, \(m\) is the mass of the object, and \(a\) is the acceleration (or in our case, deceleration). In this case, the friction force is the only force acting on the ball, and it acts opposite to its direction of motion. So, we can write: \[ a = \frac{F_{friction}}{m} \] Plugging in our values, we get: \[ a = \frac{25 \mathrm{~N}}{5 \mathrm{~kg}} = 5 \mathrm{~m/s^2} \]

Step 3: Find the Time to Stop

Now, we will find the time it takes for the ball to stop due to the deceleration caused by friction. Using the formula for uniformly accelerated motion, we have: \[ v_f = v_i - a \cdot t \] Where \(v_f\) is the final velocity (0 m/s in this case, as the ball stops), \(v_i\) is the initial velocity (10 m/s), \(a\) is the deceleration (5 m/s^2), and \(t\) is the time. Solving for \(t\), we get: \[ t = \frac{v_i - v_f}{a} = \frac{10 \mathrm{~m/s} - 0 \mathrm{~m/s}}{5 \mathrm{~m/s^2}} = 2 \mathrm{~s} \]

Step 4: Calculate the Distance Traveled

Finally, we will find the distance the ball travels before stopping using the formula for uniformly accelerated motion: \[ d = v_i \cdot t - \frac{1}{2} \cdot a \cdot t^2 \] Plugging in the values we have found, we get: \[ d = 10 \mathrm{~m/s} \cdot 2 \mathrm{~s} - \frac{1}{2} \cdot 5 \mathrm{~m/s^2} \cdot (2 \mathrm{~s})^2 \] \[ d = 20 \mathrm{~m} - 10 \mathrm{~m} = 10 \mathrm{~m} \] The ball will describe a distance of \(10 \mathrm{~m}\) before stopping. Hence, the correct answer is (D).

Key Concepts

Coefficient of Friction

The coefficient of friction is a key concept in understanding how objects move when in contact with a surface. It is a dimensionless number that represents the ratio of the force of friction between two bodies and the force pressing them together. The greater the coefficient, the more resistance one surface has when moving over another.
The formula for the frictional force is:

• \[ F_{friction} = \mu \cdot F_{normal} \]
• Where \( \mu \,\) represents the coefficient of friction and \( F_{normal} \,\) is the normal force.

In our exercise, the coefficient of friction is \( \frac{1}{2} \,\), which tells us that the frictional force is half of the normal force acting on the ball. Understanding this coefficient allows us to calculate how quickly an object will slow down or stop due to the frictional force.

Newton's Second Law

Newton's Second Law of Motion forms the basis for calculating the forces involved when objects move. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration, formulated as:

• \[ F = m \cdot a \]

This law relates directly to how we calculate the deceleration of the ball in our exercise. The frictional force is the sole force causing the ball to slow down in this scenario, and by using the equation \( F = m \cdot a \,\), we solve for \( a \,\), which is the deceleration.

• A higher frictional force would lead to higher deceleration. Conversely, if the mass of the ball was greater but the frictional force remained the same, the deceleration would be smaller.

This relationship is essential for understanding how different forces and masses affect an object's motion.

Uniformly Accelerated Motion

Uniformly accelerated motion refers to motion where the acceleration is constant over time. This is important in our exercise as it means the ball is decelerating at a consistent rate until it stops. Key equations of uniformly accelerated motion allow us to find the distance covered and the time taken to stop:

• Final velocity formula: \( v_f = v_i + a \cdot t \,\) (here \( v_f = 0 \,\) as the ball stops)
• Distance formula: \[ d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 \]

These equations give us a complete view of how the ball's velocity changes over time and help us calculate the exact distance it travels before coming to a halt. The understanding of uniformly accelerated motion is crucial for solving numerous physics problems involving constant acceleration or deceleration.

Kinetics

Kinetics is the study of the forces that cause motion, which is a central theme in our exercise. It involves analyzing the effect of forces like friction on the movement of objects, in this context, the ball. By understanding kinetics, we can predict and describe how objects will move under different forces.
Here, kinetics helps us:

• Calculate the impact of frictional forces on the ball's motion.
• Connect the concepts like Newton's Second Law and coefficient of friction to real-world scenarios.

Simply put, kinetics provides the tools needed to determine how fast an object stops or moves with given force values. In practice, this allows us to make educated predictions about motion in varied conditions, making it a vital area of study in physics and engineering alike.