Verified Solution • 2025–26 Session
hard
QJEEM_2024_2: JEE Main 2024 JEEM_2024_2 Solution: For the block shown, F1 is the minimum force required to move…
For the block shown, F1 is the minimum force required to move block upwards and F2 is the minimum force required to prevent it from slipping, find F1 - F2. (μ = 0.1)
Multiple Choice Options
- A 50 N
- B 5 N
- C 25 N
- D 2 N
Step-by-Step Verified Solution
f_k = µmg cos θ => f_k = 0.1 × 50 × (3/5) => f_k = 2.5 N; F1 = mg sin θ + f_k => F1 = 25 + 2.5 => F1 = 27.5 N; F2 = mg sin θ - f_k => F2 = 25 - 2.5 => F2 = 22.5 N; F1 - F2 = 5 N