Question

A person pushes a box on a rough horizontal plateform 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?

Answer 1

5250 J

Answer 2

The work done by a variable force is given by the area under the Force-Displacement (F-x) graph. The total movement of the box can be divided into two parts:

Part 1: Constant Force

• Force applied, $F_1 = 200 \, \text{N}$
• Distance moved, $x_1 = 15 \, \text{m}$

The work done in this part ($W_1$) is calculated as: $W_1 = F_1 \times x_1$ $W_1 = 200 \, \text{N} \times 15 \, \text{m}$ $W_1 = 3000 \, \text{J}$

Part 2: Linearly Varying Force

• The force starts at $200 \, \text{N}$ (at $x=15 \, \text{m}$) and reduces linearly to $100 \, \text{N}$ (at $x=30 \, \text{m}$).
• The distance moved in this part, $x_2 = \text{Total distance} - x_1 = 30 \, \text{m} - 15 \, \text{m} = 15 \, \text{m}$.

The force-displacement graph for this part is a straight line, forming a trapezoid. The work done ($W_2$) is the area of this trapezoid. The formula for the area of a trapezoid is $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$. Here, the parallel sides are the initial and final forces in this segment ($200 \, \text{N}$ and $100 \, \text{N}$), and the height is the distance moved ($15 \, \text{m}$).

$W_2 = \frac{1}{2} \times (F_{\text{initial}} + F_{\text{final}}) \times \Delta x$ $W_2 = \frac{1}{2} \times (200 \, \text{N} + 100 \, \text{N}) \times 15 \, \text{m}$ $W_2 = \frac{1}{2} \times (300 \, \text{N}) \times 15 \, \text{m}$ $W_2 = 150 \, \text{N} \times 15 \, \text{m}$ $W_2 = 2250 \, \text{J}$

Total Work Done The total work done by the person ($W_{\text{total}}$) is the sum of the work done in both parts: $W_{\text{total}} = W_1 + W_2$ $W_{\text{total}} = 3000 \, \text{J} + 2250 \, \text{J}$ $W_{\text{total}} = 5250 \, \text{J}$

The force-displacement graph can be visualized as: