A variable force F acts on a body moving along a straight horizontal path. The force incre… — JAMB UTME

A variable force $F$ acts on a body moving along a straight horizontal path. The force increases linearly from $0\text{ N}$ at position $x = 0\text{ m}$ to $20\text{ N}$ at $x = 4\text{ m}$ , remains constant at $20\text{ N}$ from $x = 4\text{ m}$ to $x = 7\text{ m}$ , and then decreases linearly back to $0\text{ N}$ at $x = 10\text{ m}$ . What is the total work done by the force in Joules over the $10\text{ m}$ displacement?

Answer: 130 J

Answer

130 J

The work done by a variable force is equal to the total area under the force-displacement (

F

x

) graph. The region under the graph forms a trapezoid bounded by parallel sides of length

10\text{ m}

(total displacement) and

3\text{ m}

(constant force interval), with a height of

20\text{ N}

. Calculating the area yields

W = \frac{1}{2}(10 + 3) \times 20 = 130\text{ J}

Step-by-Step Solution

Relate work done to the force-displacement graph

Work done

W

equals the total area under the

F

x

graph between

x = 0\text{ m}

and

x = 10\text{ m}

By definition,

W = \int F \, dx

, which corresponds to the geometric area under the force-displacement curve.

Calculate the geometric area under each section of the graph

First section (

0

4\text{ m}

): Triangle area =

\frac{1}{2} \times 4 \times 20 = 40\text{ J}

. Second section (

4

7\text{ m}

): Rectangle area =

(7 - 4) \times 20 = 60\text{ J}

. Third section (

7

10\text{ m}

): Triangle area =

\frac{1}{2} \times (10 - 7) \times 20 = 30\text{ J}

Breaking down a complex piecewise curve into simple geometric shapes allows straightforward area evaluation without calculus.

Sum the areas of all sections

Total Work

W = 40\text{ J} + 60\text{ J} + 30\text{ J} = 130\text{ J}

The total work done is the scalar sum of the work done across each contiguous segment of displacement.

Key Concept

Work Done by a Variable Force (Area under Force-Displacement Graph)

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