Problem 60 - Master Resource Book in JEE Main Physics

Question

A machine which is \(75 \%\) efficient uses \(12 \mathrm{~J}\) of energy in lifting up a \(1 \mathrm{~kg}\) mass through a certain distance. The mass is then allowed to fall through that distance. The velocity of the ball at the end of its fall is (a) \(\sqrt{24} \mathrm{~ms}^{-1}\) (b) \(\sqrt{32} \mathrm{~ms}^{-1}\) (c) \(\sqrt{18} \mathrm{~ms}^{-1}\) (d) \(3 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The velocity of the ball at the end of its fall is \(\sqrt{18} \mathrm{~ms}^{-1}\), so (c) is correct.

Step 1: Understanding Machine Efficiency

The machine is 75% efficient, which means only 75% of the energy supplied is used to do useful work such as lifting the mass. The energy used effectively is calculated by multiplying the total energy supplied (12 J) by the efficiency (0.75).

Step 2: Calculating Effective Work Done

Calculate the effective work done in lifting the mass. This is given by \( \text{Effective Work} = 0.75 \times 12 \ \text{J} = 9 \ \text{J} \). This effective work done equals the gravitational potential energy gained by the mass when lifted.

Step 3: Relating Potential Energy to Velocity

As the mass falls, the gravitational potential energy is converted into kinetic energy. The potential energy at the top is equal to the kinetic energy at the bottom. Hence, set the potential energy equal to the kinetic energy: \( mgh = \frac{1}{2}mv^2 \), where \( m = 1 \ \text{kg} \).

Step 4: Simplifying the Energy Equation

Since \( m \) cancels out from both sides, we have \( gh = \frac{v^2}{2} \). Substitute the potential energy \( mgh = 9 \ \text{J} \) into the kinetic energy equation \( \frac{1}{2}mv^2 = 9 \ \text{J} \) and solve for \( v^2 \).

Step 5: Solving for Velocity

By equating \( mgh = \frac{1}{2}mv^2 \) and knowing \( mgh = 9 \ \text{J} \), we have \( \frac{1}{2}v^2 = 9 \). Thus, \( v^2 = 18 \), giving \( v = \sqrt{18} \ \text{ms}^{-1} \).

Key Concepts

Machine Efficiency

Machine efficiency is a measure of how well a machine converts input energy into useful output work. It is expressed as a percentage. For instance, if a machine is 75% efficient, like in this exercise, it means that only 75% of the input energy is doing productive work while the rest is lost, usually as heat or sound.

In the given exercise, the machine uses a total of 12 joules of energy to lift a mass. However, because its efficiency is only 75%, not all this energy is used for lifting. To find out how much energy is effectively used, we multiply the total energy input by the efficiency:

• Effective energy = 12 J × 0.75 = 9 J.

This value indicates the amount of energy successfully used to lift the mass, which translates into an energy form known as potential energy. Understanding machine efficiency helps us appreciate how energy transformations are never 100% efficient in real-world scenarios.

Potential Energy

Potential energy refers to the energy stored in an object due to its position or state. In the context of our exercise, it is gravitational potential energy, which is gained when a mass is lifted to a certain height against gravity.

The formula for gravitational potential energy is given by:

• PE = mgh,

where:

• m is the mass,
• g is the acceleration due to gravity (approximately 9.8 m/s²),
• h is the height.

In this instance, the effective work converted to potential energy is 9 J, which implies that the potential energy at the lift height is 9 J. As this energy is stored due to being lifted, it becomes crucial when the mass begins to fall, turning back into kinetic energy.

Thus, potential energy serves as a temporary energy store, waiting to transform back into kinetic energy when the opportunity arises.

Kinetic Energy

Kinetic energy is the energy a body possesses due to its motion. In simple terms, any object that is moving has kinetic energy, which depends on its mass and velocity, described mathematically by the equation:

• KE = \( \frac{1}{2}mv^2 \).

In our exercise, as the mass falls, the potential energy gained when lifted is converted into kinetic energy. Since no external forces like friction are mentioned, and assuming an ideal situation, all potential energy transforms into kinetic energy.

Given that the effective potential energy stored was 9 J, the kinetic energy when the mass finishes falling is also 9 J. Setting these equal due to energy conservation gives:

• \( \frac{1}{2}mv^2 = 9 \, J \)

Rearranging and substituting for mass (1 kg) allows us to calculate the velocity of the mass when it has fallen through the distance:

• \( v^2 = 18 \), yielding \( v = \sqrt{18} \) ms^-1.

This demonstrates how energy conservation allows calculations between various energy forms and provides predictions on an object's motion from potential energy transitions.