Problem 63 - Master Resource Book in JEE Main Physics

Question

A mass of \(50 \mathrm{~kg}\) is raised through a certain height by a machine whose efficiency is \(90 \%\), the energy is \(5000 \mathrm{~J}\). If the mass is now released, its kinetic energy on hitting the ground shall be (a) \(5000 \mathrm{~J}\) (b) \(4500 \mathrm{~J}\) (c) \(4000 \mathrm{~J}\) (d) \(5500 \mathrm{~J}\)

Step-by-Step Solution

Answer

The kinetic energy is 4500 J.

Step 1: Understanding the Given Data

We know that the machine's efficiency is 90%, meaning it uses the available energy less effectively. The total input energy provided is 5000 J, and we must calculate the energy effectively used based on this efficiency.

Step 2: Calculate Useful Energy

The machine's efficiency tells us the fraction of the input energy that is converted to useful work. We calculate this as follows: \[\text{Useful Energy} = \text{Efficiency} \times \text{Total Energy} = 0.9 \times 5000 \ \text{J} = 4500 \ \text{J}\]

Step 3: Apply Energy Conservation Principles

When the mass is raised, the useful energy becomes potential energy at height. When released, all of this potential energy will be converted into kinetic energy at the point just before hitting the ground. Hence, the kinetic energy upon impact will be equal to the useful energy calculated.

Step 4: Determine the Kinetic Energy

Given the energy conservation principle, the kinetic energy when the mass hits the ground will equal the useful energy (from potential energy conversion): 4500 J.

Key Concepts

Efficiency

Have you ever wondered how efficiently machines work? In this problem, a machine works with 90% efficiency. Efficiency, simply put,
is a way to measure how effectively a machine converts input energy into useful work.
In mathematical terms, efficiency

• is expressed as a percentage.
• For our machine, it means only 90% of the input energy is used for its intended purpose.

Efficiency helps us understand both the potential and the limitations of machines.
For instance, in the provided exercise,
you are given an input energy of 5000 J. However, the machine can't use all of this energy effectively.To calculate the energy actually used for lifting,
you multiply the input energy by the efficiency percentage:\[\text{Useful Energy} = \text{Efficiency} \times \text{Total Energy}\]Plugging in the numbers,
you're given that \(0.9 \times 5000\) J = 4500 J.
This value tells us the energy effectively used to lift the mass in a productive way.

Energy conservation

Energy conservation is a key principle in physics, stating that energy cannot be created or destroyed—only transformed from one form to another.
In the scenario presented in this exercise,
the energy transitions play a crucial role.

• Initially, the available energy (4500 J) is used to raise the mass, converting it into potential energy.
• Once released, the potential energy shifts back to kinetic energy.

This seamless transfer among energy forms illustrates the energy conservation principle. Imagine the mass at the highest point having potential energy that precisely equals the useful energy from efficiency calculations,
which is 4500 J.
When the mass is free to fall, all this potential energy morphs into kinetic energy by the time it hits the ground.
This remarkable transformation from potential to kinetic energy ensures that despite different forms, the total energy retains the same value.

Potential energy

Potential energy is the stored energy an object possesses due to its position or state.
For example, raising a mass to a particular height builds potential energy.
In our exercise, the mass of 50 kg is elevated for this very reason.
The potential energy stored in the mass shows how energy is temporarily captured and ready to perform work.
As with our machine, 4500 J of energy was used to lift the mass and was stored as potential energy. To visualize how potential energy works, consider this:

• This kind of energy depends on two main factors: the height and mass of the object.
• The greater the height and mass, the more potential energy there is.

When ultimately released, the potential energy is not lost.
Instead, it is transformed back into kinetic energy, indicating efficient use and conservation of energy overall in this lifting and dropping process.