A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a dis

Q: Why can't I just use $W = F \times d$ for the whole thing?

That formula only works when the force is strictly constant throughout the entire displacement. Here, the force changes in the second half of the journey, so you must use integration or find the area under the $F-x$ graph.

Q: How do I know the second part of the graph is a straight line?

The problem explicitly states that the applied force "reduces linearly". In mathematics and physics, a linear relationship always graphs as a straight line.

Q: What is the formula for the area of a trapezium?

The area of a trapezium is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{distance between them})$. In our graph, the parallel sides are the vertical force values ($200$ N and $100$ N), and the distance between them is the displacement interval ($15$ m).

Q: Could I have solved this using integration instead of geometry?

Absolutely! You would first find the equation of the straight line for the second phase: $F(x) = 300 - \frac{20}{3}x$. Then, you would integrate this function from $x=15$ to $x=30$. However, the graphical method is usually faster and less prone to calculation errors.

Q: What if the force reduced exponentially instead of linearly?

If the force reduced exponentially, the graph would be a curve, not a straight line. In that case, you could not use the simple trapezium area formula. You would be forced to find the mathematical function $F(x)$ and use calculus (integration) to find the exact area.

Question

A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?

Options:

(A) 5250 J

(B) 2780 J

(C) 3280 J

(D) 5690 J

Sigma Percentile · Accepted Answer

### The Setup
Imagine you are tasked with pushing a heavy box across a rough warehouse floor. At first, you are full of energy. You push with a steady, constant force of $200$ N, and you manage to move the box a solid $15$ m.

But then, fatigue starts to set in. Your muscles burn, and your pushing force begins to drop. The problem states that your force *reduces linearly* with distance until it hits $100$ N at the $30$ m mark.

When a force is constant, calculating work is as simple as multiplying force by distance ($W = F 	imes d$). But what do we do when the force is constantly changing?

### The Graphical Masterstroke
While you could find the mathematical equation for the force and use calculus to integrate it, there is a much more elegant and visual way to solve this: **The Force-Displacement Graph**.

In physics, the total work done by a variable force is exactly equal to the **area under the Force vs. Displacement ($F-x$) curve**. Let's sketch this out.

### Phase 1: The Constant Push
For the first $15$ m, the force is a constant $200$ N. On our graph, this is a horizontal line from $x = 0$ to $x = 15$ at a height of $F = 200$.

The area under this part of the graph is simply a rectangle. Let's call this $A_1$.

$$A_1 = 	ext{length} 	imes 	ext{width}$$

$$A_1 = 200 	imes 15 = 3000 	ext{ J}$$

So, you did $3000$ Joules of work while you were fresh.

### Phase 2: The Fading Strength
From $x = 15$ m to $x = 30$ m, the force drops linearly from $200$ N to $100$ N. "Linearly" means this is a straight, downward-sloping line.

The area under this section forms a trapezium. Let's call this $A_2$. The formula for the area of a trapezium is half the sum of the parallel sides multiplied by the distance between them.

$$A_2 = \frac{1}{2} 	imes (F_{	ext{initial}} + F_{	ext{final}}) 	imes \Delta x$$

$$A_2 = \frac{1}{2} 	imes (200 + 100) 	imes (30 - 15)$$

$$A_2 = \frac{1}{2} 	imes 300 	imes 15 = 2250 	ext{ J}$$

Even while getting tired, you still managed to do $2250$ Joules of work.

### The Grand Total
To find the total work done over the entire $30$ m journey, we simply add the two areas together.

$$W_{	ext{total}} = A_1 + A_2$$

$$W_{	ext{total}} = 3000 + 2250 = \mathbf{5250 	ext{ J}}$$

This perfectly matches option (a).

**A Note on Integration:** You could have found the equation of the line for the second phase, which is $F(x) = 300 - \frac{20}{3}x$, and evaluated the integral $\int_{15}^{30} \left(300 - \frac{20}{3}xight) dx$. It would yield the exact same result of $2250$ J. However, recognizing the geometric shapes saves time and drastically reduces the chance of algebraic errors. Always keep an eye out for graphical shortcuts!

Select Answer:

Visualized Solution

$W = \int F d x$

$Phase 1: Constant Force$

$Phase 2: Linear Decrease$

$Total Work Done$

$Calculating A_{1}$

$Calculating A_{2}$

$Final Answer$

$Alternative: Integration$

The Sigma Insight: Work Done by Forces

The Setup Imagine you are tasked with pushing a heavy box across a rough warehouse floor

The Graphical Masterstroke

Phase 1

Phase 2

The Grand Total

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=∫Fdx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phase 1: Constant Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phase 2: Linear Decrease

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Work Done

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating A1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating A2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Alternative: Integration