Retardation of an object Sliding up the inclined plane

# Retardation of object sliding up the inclined plane: -> → equilibrium is applied for to movement N mg Fk motion mg sinθ mg cosθ Eom → Fnet = ma -mg sinθ - Fk = ma -mg sinθ - μN = ma -mg sinθ - μmg cosθ = ma -mg(sinθ + μ cosθ) = ma 0 = mg cosθ a = -g(sinθ + μ cosθ) QA block of mass 2kg is at rest on a horizontal rough surface (μ=0.5) A horizontal force F=5t is applied on the block at t=0. What is speed after 9 sec. Solution: N=mg [by equilibrium] F=5t mg f(max) = μN = 0.5 × 2 × 10 = 10 N F = 5t = 10 ⇒ t = 10/5 = 2 sec ∴ the block is still in rest till t=2 sec, after 2 sec, the block will move. Now equation of motion 5t - μmg = ma 5t - 10 = dv/dt ∫dv = ∫(5t - 10)dt v = [5t²/2 - 10t]₀⁴ v = [5(16)/2 - 10(4) - 5(0)/2 + 10(0)] = 1/2[40 - 40 - 10 + 30] = 10/2 = 5 v = 5 m/s Ans Note → For max or min. calculation calculate (Fmax) by considering limiting equilibrium (about to move object). D→ A 2kg block is lying on a rough surface. If coefficient of static friction bhw the black and ground is 0.3. What max. force can be applied on the black without moving it. Solution: EOM: Fnet = 0 F(max) mg Fsin45° - μN(mg + Fcos45°) = ma = 0 F/√2 - 0.3(9.810 + F/√2) = 0 F/√2 - 0.3(9.810) - 0.3F/√2 = 0 (0.7)F/√2 = 0.3(9.810) F = 4√2 / 0.7 F = 7.07 N Ans Q If In the figure shown coefficient of friction b/w the black and inclined plane is μ. What should be the value of F so that black slides up the plane with constant speed. \[F~cos~\theta\] Salution → limit equilibrium \[N=mg~cos~\theta+F~sin~\theta\] F \[F~\xi in\theta\] mgreso velacity ng sinde If black slides with constant accelerom F then \[a=0\] my EOM \[F_{net}=ma=0\] \[F~cos~\theta-mg~sin~\theta-F_{k}=0\] → \(F~cos~\theta-mg~sin~\theta-\mu(m~g~cos~\theta+F~sin~\theta)=\emptyset\) → \(F(cas~\theta-\mu~sin~\theta)-m~g(sin~\theta+\mu~ces~\theta)=0\) \[F=mg(sin~\theta+\mu~cos~\theta)\] Ans A