JEE Main 2023 · 1 February · morning · Single correct · +4 / −1
A block of mass 5 kg is placed at rest on a table of rough surface. Now, if a force of 30 N is applied in the direction parallel to surface of the table, the block slides through a distance of 50 m in an interval of time 10 s. Coefficient of kinetic friction is (given, g=10 ms⁻²):
- (a) 0.25
- (b) 0.75
- (c) 0.60
- (d) 0.50
Answer and solution
Answer (d) 0.50
Solution
Physics sir · 4 steps
Starting from rest, the 50 m in 10 s gives you the acceleration. Then Newton's second law hands you μ.
- Find the acceleration. The block starts from rest, so u = 0. Use s = ut + 1/2at² with s = 50 m and t = 10 s.
50 = 1/2 × a × 10² ⇒ a = 1 m s⁻²
- Forces along the table. The applied force F = 30 N pulls forward, and kinetic friction μₖ mg = μₖ × 5 × 10 = 50μₖ pulls back. So F - μₖ mg = ma.
- Solve for μ. 30 - 50μₖ = 5 × 1 = 5, so 50μₖ = 25 and μₖ = 0.5.
- So. The coefficient of kinetic friction is 0.50, option (d).
The idea to keep. When a block is pulled across a rough surface, the net force is the applied force minus μₖ mg, and that net force equals ma. Get a from kinematics first.
Answer from the NTA final answer key. Solution by PrepWiser.