A force varying linearly with the distance acts on a body as shown above, The work done on the body by the force during the first 10m of motion is
Attached Figure / Diagram
A
100J
B
150J
C
200J
D
300J
E
600J
Verified Correct Answer: B.
Physical Principles & Numerical Solution
The graph shows the force decreasing linearly from 20 N at 0 m to 0 N at 20 m.
For the first 10 m:
- Force at 0 m = 20 N
- Force at 10 m = 10 N (midway)
The area under the graph (work done) is the area of a trapezium:
\(\text{Work} = \frac{1}{2} \times (20 + 10) \times 10 = 15 \times 10 = \mathbf{150\, J}\).
Source: Myschool Past Questions Archive (Record #41643)