University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

86

Pages:

5
Author:

Jabin F.

# Problems based on friction: **Example:** Block at rest on inclined plane. [Image of a block on an inclined plane with forces labeled] For equilibrium of block: N = mg cosθ Fs = mg sinθ Let μ is coefficient of friction (Fs)max = μN = μmg cosθ Fs ≤ (Fs)max mg sinθ ≤ μmg cosθ tanθ ≤ μ → An object placed on an inclined plane can stay at rest if tanθ ≤ μ; object will slide if tanθ > μ. **2. A particle starts from rest and moves with acceleration α for time t₁. Then it moves with retardation β and finally comes to rest. If total distance moved is S, calculate max. speed attained. Calculate average speed.** **Solution:** Now equation of motion for: 1st part: v₀ = 0 + αt₁ v₀ = αt₁ v₀² = 0 + 2αx v₀² = 2αx For 2nd part: v = u + at 0 = v₀ - βt₂ t₂ = v₀/β v² = u² + 2as 0 = v₀² - 2β(S - x) v₀² = 2β(S - x) - (2) From (1) and (2): 2αx = 2β(S - x) αx = β(S - x) αx = βS - βx αx + βx = βS x(α + β) = βS x = βS/(α + β) Total time: T = t₁ + t₂ T = v₀/α + v₀/β T = v₀(α + β)/αβ From eq. (1) & (2) we get αx = β(S - x) => αx + βx = βS x = βS/(α + β) Put in eq. (1) we get v₀² = 2αβS/(α + β) v₀ = √(2αβS/(α + β)) Now avg. speed = total distance / total time = S/T = S/T = S / [v₀(α + β)/αβ] = αβS / [v₀(α + β)] = αβS / [√(2αβS/(α + β)) * (α + β)] = √(αβS/2(α + β)) 2. A ball is thrown upward with speed √(2gh) from top of a tower of height h. After what time, ball will reach the ground? Solution: The ball will first go upto a height h against gravity and come to rest at this position. So time taken in the motion, t₁ = √(h/g) Now the ball will fall down a height 2h from rest to reach ground. So time taken in the motion, t₂ = √(2h/g) Now the total time taken by the ball to reach ground is: T = t₁ + t₂ = √(h/g) + √(2h/g) T = √(h/g)(1 + √2) A particle is projected with speed u0 at 45° with horizontal. After what time, KE of particle will become 3/8 of its initial KE. (u0 = 10 m/s) Solution: Initial velocity: u = u0 = 10 m/s. At any instant velocity: v = u + at = u0 cos45°i + (u0sin45° - gt)j According to the question, the kinetic energy will be: (1/2)mu² = (3/8)(1/2)m(v²) 9u₀² = v² + (u0 - gt)² 9u₀² = u0²(cos²45° + (sin45° - gt)²) 9/8 = 1 + (1 - gt)² (1 - gt)² = 1/8 1 - gt = ±1/√8 gt = 1 ± 1/√8 t = u0/g(1 ± 1/√8) t = 10/10(1 ± 1/√8) t = 1 ± 1/√8