JEE Main 2022 · 28 June · evening · Single correct · +4 / −1
A block of mass 2 kg moving on a horizontal surface with speed of 4 ms⁻¹ enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12 Nm⁻¹. The speed of the block as it just crosses the rough surface will be :
- (a) zero
- (b) 1.5 ms⁻¹
- (c) 2.0 ms⁻¹
- (d) 2.5 ms⁻¹
Answer and solution
Answer (c) 2.0 ms⁻¹
Solution
Physics sir · 4 steps
The force changes with x, so integrate it to get the work, then let the work-energy theorem give the speed.
- Starting energy. Before the rough patch the kinetic energy is 1/2 × 2 × 4² = 16 J.
- Work by friction. The retarding force F = -kx grows as the block moves, so you cannot use force times distance. Integrate it over the rough patch only, from x = 0.5 m to x = 1.5 m.
W = -∫_(0.5)^(1.5) 12x dx = -6(1.5² - 0.5²) = -6 × 2 = -12 J
- Work-energy theorem. Final kinetic energy is 16 - 12 = 4 J. So 1/2 × 2 × v² = 4, which gives v² = 4.
- So. The block leaves the rough surface at v = 2.0 m s⁻¹, which is option (c).
The idea to keep. When a force depends on position, work is ∫ F dx between the actual start and end points. Then change in kinetic energy equals that work.
Answer from the NTA final answer key. Solution by PrepWiser.