A body of mass 0.5 kg travels in a straight line with velocity $v = a x^{3/2}$ where $a = 5$ m$^{-1/2}$s$^{-1}$. The work done by the net force during its displacement from $x = 0$ to $x = 2$ m is (a) 1.5 J (b) 50 J (c) 10 J (d) 100 J

> [!TLDR] > The work done by the net force on a body is equal to the change in its kinetic energy. By calculating the kinetic energy at the initial and final positions, we find the work done to be $\boxed{50 \text{ J}}$. When a problem asks for the work done by the *net force* and provides information about how velocity changes with position, the Work-Energy Theorem is almost always the most straightforward and efficient approach. This is because the theorem directly relates the net work done to the change in the body's kinetic energy, bypassing the need to first find the net force and then integrate it over the displacement. > [!FORMULA] > The Work-Energy Theorem states that the work done by the net force ($W_{net}$) on an object is equal to the change in its kinetic energy ($\Delta KE$): > $$W_{net} = \Delta KE = KE_{final} - KE_{initial} = \frac{1}{2}mv_{final}^2 - \frac{1}{2}mv_{initial}^2$$ Let's apply this theorem step-by-step: 1. **Identify the given information:** * Mass of the body, $m = 0.5$ kg. * Velocity of the body as a function of position, $v = a x^{3/2}$. * Value of the constant, $a = 5$ m$^{-1/2}$s$^{-1}$. * Initial position, $x_1 = 0$ m. * Final position, $x_2 = 2$ m. 2. **Calculate the initial velocity and initial kinetic energy:** At the initial position $x_1 = 0$ m, the velocity $v_1$ is: $v_1 = a (0)^{3/2} = 0$ m/s The initial kinetic energy $KE_1$ is: $KE_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} (0.5 \text{ kg}) (0 \text{ m/s})^2 = 0$ J 3. **Calculate the final velocity and final kinetic energy:** At the final position $x_2 = 2$ m, the velocity $v_2$ is: $v_2 = a (2)^{3/2}$ Substitute the value of $a = 5$: …

A body of mass 0.5 kg travels in a straight line with velocity v = a x3/2 where a = 5 m-1/2s⁻¹. The work done

NCERT Exemplar · Q10

v = a x^{3/2}

(a) 1.5 J

(b) 50 J

(d) 100 J

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't) …

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The work done by the net force on a body is equal to the change in its kinetic energy. By calculating the kinetic energy at the initial and final positions, we find the work done to be $\boxed{50 \text{ J}}$ .

When a problem asks for the work done by the net force and provides information about how velocity changes with position, the Work-Energy Theorem is almost always the most straightforward and efficient approach. This is because the theorem directly relates the net work done to the change in the body's kinetic energy, bypassing the need to first find the net force and then integrate it over the displacement.

W_{net}

Let's apply this theorem step-by-step:

Identify the given information:
- Mass of the body, $m = 0.5$ kg.
- Velocity of the body as a function of position, $v = a x^{3/2}$ .
- Value of the constant, $a = 5$ m $^{-1/2}$ s $^{-1}$ .
- Initial position, $x_1 = 0$ m.
- Final position, $x_2 = 2$ m.
Calculate the initial velocity and initial kinetic energy:

At the initial position $x_1 = 0$ m, the velocity $v_1$ is:

$v_1 = a (0)^{3/2} = 0$ m/s

The initial kinetic energy $KE_1$ is:

$KE_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} (0.5 \text{ kg}) (0 \text{ m/s})^2 = 0$ J
Calculate the final velocity and final kinetic energy:

At the final position $x_2 = 2$ m, the velocity $v_2$ is:

$v_2 = a (2)^{3/2}$

Substitute the value of $a = 5$ : …

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Q.A body of mass 0.5 kg travels in a straight line with velocity $v = a x^{3/2}$ where $a = 5$ m $^{-1/2}$ s $^{-1}$ . The work done by the net force during its displacement from $x = 0$ to $x = 2$ m is

Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

The Intuition First

The Precise Statement

Why "Net" Work Matters

A Simple Derivation (for constant force)

What It Tells You (and What It Doesn't) …

Concept: Work-Energy Theorem

Q.A body of mass 0.5 kg travels in a straight line with velocity v=ax3/2v = a x^{3/2}v=ax3/2 where a=5a = 5a=5 m−1/2^{-1/2}−1/2s−1^{-1}−1. The work done by the net force during its displacement from x=0x = 0x=0 to x=2x = 2x=2 m is

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Q.A body of mass 0.5 kg travels in a straight line with velocity $v = a x^{3/2}$ where $a = 5$ m $^{-1/2}$ s $^{-1}$ . The work done by the net force during its displacement from $x = 0$ to $x = 2$ m is

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