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Question

What will be the value of the kinetic energy (EK) of a moving body with mass m, if its speed is doubled from v to 2v?

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is

4 E k

Understanding Kinetic Energy and Speed

Kinetic energy is the energy an object possesses due to its motion. The amount of kinetic energy a body has depends on its mass and its speed. The standard formula for kinetic energy \(E_k\) of a body with mass \(m\) and speed \(v\) is given by:

\( E_k = \frac{1}{2}mv^2 \)

This formula shows that kinetic energy is directly proportional to the mass and the square of the speed. This means if you change either the mass or the speed, the kinetic energy will change.

Calculating Change in Kinetic Energy when Speed is Doubled

Let's consider a body with mass \(m\) moving with an initial speed \(v\). The initial kinetic energy, which we can call \(E_{k1}\), is:

\( E_{k1} = \frac{1}{2}mv^2 \)

Now, the question states that the speed is doubled. So, the new speed, let's call it \(v_2\), is \(2v\). The mass \(m\) remains the same. The new kinetic energy, \(E_{k2}\), will be:

\( E_{k2} = \frac{1}{2}m(v_2)^2 \)

Substitute the new speed \(v_2 = 2v\) into the formula:

\( E_{k2} = \frac{1}{2}m(2v)^2 \)

Now, we need to calculate \((2v)^2\). This is \((2v) \times (2v) = 4v^2\).

So, the expression for \(E_{k2}\) becomes:

\( E_{k2} = \frac{1}{2}m(4v^2) \)

We can rearrange this expression:

\( E_{k2} = 4 \times \left(\frac{1}{2}mv^2\right) \)

Notice that the expression inside the parentheses, \(\frac{1}{2}mv^2\), is exactly the formula for the initial kinetic energy \(E_{k1}\). Therefore, we can write:

\( E_{k2} = 4 E_{k1} \)

This shows that when the speed of the body is doubled, its kinetic energy becomes four times its initial value.

Comparing Options for Kinetic Energy Change

We found that the new kinetic energy \(E_{k2}\) is \(4\) times the initial kinetic energy \(E_{k1}\). Let's compare this with the given options, assuming \(E_k\) in the options represents the initial kinetic energy \(E_{k1}\).

Option Description Matches Calculation?
\( \frac{1}{2} E_k \) The new kinetic energy is half the original. No
\( 4 E_k \) The new kinetic energy is four times the original. Yes
There will be no change in \( E_k \) The kinetic energy remains the same. No
\( 2 E_k \) The new kinetic energy is double the original. No

Our calculation \(E_{k2} = 4 E_{k1}\) directly matches the option \(4 E_k\), assuming \(E_k\) in the options represents the initial kinetic energy.

Revision Table: Kinetic Energy Concepts

Concept Formula Relationship with Speed
Kinetic Energy \( (E_k) \) \( E_k = \frac{1}{2}mv^2 \) Directly proportional to the square of speed (\( E_k \propto v^2 \)).
Mass \( (m) \) N/A Directly proportional to kinetic energy (\( E_k \propto m \)) for a constant speed.
Speed \( (v) \) N/A If speed is doubled, kinetic energy is quadrupled (\( (2v)^2 = 4v^2 \)).

Additional Information on Energy and Motion

Kinetic energy is a scalar quantity, meaning it only has magnitude and no direction. It is measured in joules (J) in the SI system.

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. Work is the transfer of energy by mechanical means.

\( W_{net} = \Delta E_k = E_{k, final} - E_{k, initial} \)

Potential energy is another form of mechanical energy, related to the position or state of an object. The total mechanical energy is the sum of kinetic and potential energy (\( E_{total} = E_k + E_p \)). In the absence of non-conservative forces like friction, the total mechanical energy is conserved.

Understanding the relationship between speed and kinetic energy is fundamental in physics and is applied in many areas, such as analyzing collisions, motion, and energy transformations.

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