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

2024
Views:

411

Pages:

5
Author:

Jabin F.

# Work-Energy Theorem (always valid) 1st Statement: Total work done = Change in K.E. (Work done by all forces) 2nd Statement: Work done by non-conservative force = Change in total energy Work done by conservative force = -ve change in potential energy Note: Potential energy is used for all conservative forces. W_total = KE_f - KE_i W_nc + W_c = KE_f - KE_i W_c = ∫Fdx = -∫dU W_c = U_i - U_f W_nc = (KE_f + U_f) - (KE_i + U_i) One particle is very light, so the another particle will not move. m is falling on M, so only force b/w these particles at any instant of time will be considered. Fnet = GmMm/(9R-x)² = ma (9R-x)² = GM/m d²x/dt² is 2nd order differential equation, hard to solve. So we first find out v. GmMm/(9R-x)² = m v dv/dx GMdx/(9R-x)² = v dv On integrating, GM ∫(1/(9R-x)²) dx = ∫v dv GM/(9R-x) = v²/2 v² = 2GM/(9R-x) v = √(2GM/(9R-x)) Now let x = 9R sin²θ → dx = 18R sinθ cosθ dθ when x = 0 → θ = 0 when x = 9R → θ = π/2 GM ∫(1/(9R-x)²) dx = ∫v dv GM ∫(1/(9R-9Rsin²θ)²) 18R sinθ cosθ dθ = ∫v dv GM/R ∫(1/(cos²θ)²) sinθ cosθ dθ = v²/2 GM/R ∫sec²θ dθ = v²/2 GM/R tanθ = v²/2 when θ = π/4 GM/R tan(π/4) = v²/2 v² = GM/R v = √(GM/R)