Problem 11 - Master Resource Book in JEE Main Physics

Question

A ball is released from the top of a tower. The ratio of work done by force of gravity in Ist second, 2nd second and 3 rd second of the motion of ball is (a) \(1: 2: 3\) (b) \(1: 4: 16\) (c) \(1: 3: 5\) (d) \(1: 9: 25\)

Step-by-Step Solution

Answer

(c) 1: 3: 5

Step 1: Understanding Work Done by Gravity

When a ball is dropped from a height, the work done by gravity in any time interval is equal to the force (which is the weight of the ball) multiplied by the distance it falls during that time interval. The distance fallen in any 'n-th' second can be calculated using the equation: \[ s_n = u + \frac{1}{2} g (2n - 1) \] where \( u = 0 \) (initial velocity), \( g \) is the acceleration due to gravity, and \( n \) is the time interval in seconds.

Step 2: Calculating Distance Fallen in Each Second

For each second, we calculate the distance fallen: - First second: \( s_1 = 0 + \frac{1}{2} \times g \times (2 \times 1 - 1) = \frac{g}{2} \)- Second second: \( s_2 = 0 + \frac{1}{2} \times g \times (2 \times 2 - 1) = \frac{3g}{2} \)- Third second: \( s_3 = 0 + \frac{1}{2} \times g \times (2 \times 3 - 1) = \frac{5g}{2} \)

Step 3: Determining Work Done in Each Second

The work done by gravity is given by \( W = m \cdot g \cdot s \) Using the distances:- First second: \( W_1 = m \cdot g \cdot \frac{g}{2} = \frac{m g^2}{2} \)- Second second: \( W_2 = m \cdot g \cdot \frac{3g}{2} = \frac{3m g^2}{2} \)- Third second: \( W_3 = m \cdot g \cdot \frac{5g}{2} = \frac{5m g^2}{2} \)

Step 4: Calculating the Ratio of the Work Done

We now calculate the ratio of work done for each second:- Ratio \( W_1:W_2:W_3 = \frac{m g^2}{2}:\frac{3m g^2}{2}:\frac{5m g^2}{2} \)- Simplifying, we get: \( 1:3:5 \)

Step 5: Comparing with Given Options

The calculated ratio of the work done by gravity in the first, second, and third seconds is \( 1:3:5 \), which corresponds to option (c).

Key Concepts

Acceleration Due to Gravity

The acceleration due to gravity, denoted as \( g \), is a key concept when dealing with objects in free fall, like the ball released from a tower in our example. On Earth, this acceleration has a constant value of approximately \( 9.81 \, \text{m/s}^2 \). This means that any object falling freely under the influence of gravity will accelerate at this rate, regardless of its mass.

When an object is released from rest, its initial velocity \( u \) is zero. In such a scenario, the velocity of the object at any moment can be calculated using the formula \( v = u + gt \), where \( t \) is the timeElapsed since the object started falling. This concept simplifies calculations, as shown in the solution where the initial velocity is zero which simplifies our equations for distance traveled.

Understanding \( g \) is crucial for calculating how fast an object speeds up as it falls, as well as the distance it covers over time. This applies to a wide range of practical and theoretical problems involving motion under gravity.

Kinematics

Kinematics is the branch of physics that describes the motion of objects without considering the forces that cause this motion. For the problem at hand, we use kinematic equations to determine how far the ball travels each second as it falls from the tower.

Since the ball falls freely from rest, our kinematic equation for distance (\( s_n \)) covered during the \( n \)-th second is highly useful:\[ s_n = u + \frac{1}{2} g (2n - 1) \] - Here, \( u \) is the initial velocity (zero in this case), \( g \) is the acceleration due to gravity, and \( n \) is the second under consideration.

From the given problem:

• First second: the ball falls a distance of \( \frac{g}{2} \).
• Second second: it covers an additional \( \frac{3g}{2} \).
• Third second: it falls \( \frac{5g}{2} \).

The calculations provide us step-by-step distances that help in further determining the work done by gravity, a direct consequence of kinetic analysis of the motion.

Motion Under Gravity

Motion under gravity refers to the way objects move when the only force acting on them is gravity. It's an exciting aspect of physics that describes scenarios similar to our exercise, where a ball is dropped from a great height.

In such motions, objects experience constant acceleration (which is acceleration due to gravity) and their paths are predictable using equations of motion. Importantly, all objects, regardless of their mass, fall at the same rate in a vacuum: approximately \( 9.81 \, \text{m/s}^2 \).
When examining an object's motion over specific intervals, such as different seconds of a fall, we use the concept of *displacement per time*. As demonstrated:

• During the first second, it displaces by \( \frac{g}{2} \).
• In the second, \( \frac{3g}{2} \) is covered.
• Finally, an additional \( \frac{5g}{2} \) is traversed in the third second.

This regular increase correlates to constant acceleration effect, providing insights into how work done by gravity affects motion.