Limiting friction force is, fmax=μ(mgcosθ)
=0.1×250×3
=2.53N
For first case:
We can write, F1=mgsinθ+fmax
=25+2.53
For second case:
F2=mgsinθ−fmax
=25−2.53
∴F1−F2=53N
A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F_1 is the force required to just move the block up the…
JEE Main 2024 — Physics Mechanics
2024mcqhard
A block of mass 5kg is placed on a rough inclined surface as shown in the figure. If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent the block from sliding down, then the value of ∣F1∣−∣F2∣ is: [Use g=10ms−2]
Official previous-year question
Held on 31 Jan 2024 · Verified 6 Jul 2026.
Options
- A
253N
- B
53N
- C
253N
- D
10N
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