A particle which is experiencing a force, is given by F = 3i - 12j, undergoes a displacement of d =

Q: Why do we completely ignore the $-12\hat{\mathbf{j}}$ component of the force?

Because work is defined as the dot product of force and displacement. Since the displacement is purely in the x-direction ($4\hat{\mathbf{i}}$), the y-component of the displacement is zero. The dot product multiplies corresponding components, so $-12 \times 0 = 0$. Physically, a force perpendicular to the direction of motion does zero work.

Q: What if the work done turned out to be negative?

If the work done was negative, it would mean the force was acting in the opposite direction of the motion, effectively acting as a brake. In that case, the final kinetic energy would be less than the initial kinetic energy.

Q: Can we solve this using Newton's equations of motion ($v^2 = u^2 + 2as$)?

Technically yes, but it would be much harder. You would need the mass of the particle to find acceleration ($\mathbf{a} = \mathbf{F}/m$) and initial velocity. The Work-Energy Theorem is a powerful shortcut that bypasses the need for mass and time entirely.

Q: Is the force considered constant in this problem?

Yes. The force vector $\mathbf{F} = 3\hat{\mathbf{i}} - 12\hat{\mathbf{j}}$ does not contain any variables like $x$, $y$, or $t$. This means it doesn't change with position or time, allowing us to use the simple dot product formula $W = \mathbf{F} \cdot \mathbf{d}$ instead of integration.

Q: What does the dot product physically represent here?

The dot product extracts the component of the force that is parallel to the displacement and multiplies it by the magnitude of the displacement. It filters out any perpendicular forces that don't contribute to changing the object's speed.

Question

A particle which is experiencing a force, is given by $\mathbf{F} = 3\hat{\mathbf{i}} - 12\hat{\mathbf{j}}$, undergoes a displacement of $\mathbf{d} = 4\hat{\mathbf{i}}$. If the particle had a kinetic energy of $3 	ext{ J}$ at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?

Options:

(A) $9 	ext{ J}$

(B) $15 	ext{ J}$

(C) $12 	ext{ J}$

(D) $10 	ext{ J}$

Sigma Percentile · Accepted Answer

Imagine a particle as a bank account, but instead of money, it stores **Kinetic Energy**. When a force acts on this particle and moves it, the force is essentially making a deposit or a withdrawal from this energy account. This beautiful relationship is governed by the **Work-Energy Theorem**.

### Visualizing the Vectors

In our scenario, the particle starts with an initial energy balance of $K_i = 3 	ext{ J}$.

It experiences a constant force given in vector form as:

$$ \mathbf{F} = 3\hat{\mathbf{i}} - 12\hat{\mathbf{j}} 	ext{ N} $$

This force pushes the particle, causing a displacement purely along the x-axis:

$$ \mathbf{d} = 4\hat{\mathbf{i}} 	ext{ m} $$

Notice that while the force is pulling the particle downwards (due to the $-12\hat{\mathbf{j}}$ component), the particle only moves horizontally.

### The Dot Product Magic

To find out how much energy the force transfers to the particle, we calculate the **Work Done ($W$)**. For a constant force, work is the dot product of the force and displacement vectors:

$$ W = \mathbf{F} \cdot \mathbf{d} $$

Let's substitute our vectors:

$$ W = (3\hat{\mathbf{i}} - 12\hat{\mathbf{j}}) \cdot (4\hat{\mathbf{i}} + 0\hat{\mathbf{j}}) $$

The dot product multiplies corresponding components. The y-component of the displacement is zero, meaning the downward pull of the force does absolutely no work! It transfers zero energy because there is no motion in that direction.

$$ W = (3 	imes 4) + (-12 	imes 0) = 12 	ext{ J} $$

The force has successfully deposited $12 	ext{ J}$ of energy into our particle's account.

### The Work-Energy Theorem

Now, we invoke the **Work-Energy Theorem**, which states that the net work done on an object equals its change in kinetic energy:

$$ W = \Delta K = K_f - K_i $$

We know the work done ($12 	ext{ J}$) and the initial kinetic energy ($3 	ext{ J}$). Let's plug them in:

$$ 12 = K_f - 3 $$

Solving for the final kinetic energy $K_f$:

$$ K_f = 12 + 3 = 15 	ext{ J} $$

The particle ends up with $15 	ext{ J}$ of kinetic energy. By breaking the problem down into vector components and applying the fundamental theorem of work and energy, we arrive at the solution elegantly and effortlessly.

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Work Done Formula}

\text{Substituting Vectors}

\text{Calculating Work Done}

\text{Work-Energy Theorem}

\text{Substituting Values}

\text{Final Kinetic Energy}

The Sigma Insight: Work Done by Forces

Visualizing the Vectors

The Dot Product Magic

The Work-Energy Theorem

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