General Science
ID: #5038 | Difficulty: hard
AI Quick Answer:
The correct option for this defence exam question is (A) which is 8349.6 J. Detailed proof and explanation are supplied below.
A soldier pulls a wooden crate of mass 71 kg over a horizontal distance of 40 m. If the coefficient of friction is 0.3, find the work done against friction (Take g = 9.8 m/s²).
Key Takeaways for NDA & CDS Candidates
- Ensure you recall the standard formulas for General Science during UPSC exam preparation.
- Eliminating obviously incorrect choices based on dimensional analysis or limits saves valuable seconds.
- Expert tip: Written defence mock tests should be practiced weekly with strict negative marking rules.
Correct Answer: (A)
8349.6 J
Step-by-Step Explanation
Work done against friction is given by:
W = f \times d
Where:
f = \mu \times R = \mu \times m \times g
d = 40 m
\mu = 0.3
m = 71 kg
g = 9.8 m/s²
f = 0.3 \times 71 \times 9.8 = 208.74 N
W = 208.74 \times 40 = 8349.6 Joules.
W = f \times d
Where:
f = \mu \times R = \mu \times m \times g
d = 40 m
\mu = 0.3
m = 71 kg
g = 9.8 m/s²
f = 0.3 \times 71 \times 9.8 = 208.74 N
W = 208.74 \times 40 = 8349.6 Joules.
| Question Metadata Summary | |
|---|---|
| Exam Category | UPSC NDA / CDS / AFCAT / Agniveer |
| Subject Unit | CDS Questions |
| Sub-topic / Module | General Science |
| Correct Option | Option (A) |
| Target Year | UPSC Defence Exams 2026 |