4r (u4, cos10+sin 10)=514.32(N/m) b)Down the slop Apply Newton s second law of motion, we have F-f+ mg sin10°=0 N- mg cosl0°=0 f=uN F=k△x △x="(cosl0°-sin10°)=0.014(m) II. Give the solutions of the following problems 1. A 1.50 kg mass on a horizontal frictionless surface is attached to a horizontal spring with spring constant k=200 N/m. The mass is in equilibrium at x=0 m. The mass is released when t=0 s at coordinate x=0. 100 m with a velocity(2.00 m/s)i (a) Find the constants A, o and o in x(()=Acos(ot +o) b) Find the period Tof the oscillation. (c) Determine the maximum speed of the oscillation and the magnitude of the maximum acceleration (d) Plot x(o during the time interval 1=0 s and /=2T. Solution (a)From the problem, 0=1 =133. 3 N/kg.m. so substitute A Vm v1.5 0.100 m and 1=2.00 m/s to x(o=Acos(at +o), we have vo =2=-Aasin=-133.3Asing 0.12+( =0.(m) Solving the equations, we have 133.3 丌 e=am-13x33 Thinking of the direction of the velocity of the mass at FO s, we get( cos10 + sin10 ) = 514.32 (N/m) ∆ ⇒ = o o k x mg k µ (b) Down the slop Apply Newton’s second law of motion, we have ( cos10 sin10 ) 0.014 (m) cos10 0 sin10 0 ⇒ ∆ = − = ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = ∆ ′ = − = − + = o o o o k k k mg x F k x f N N mg F f mg µ µ III. Give the Solutions of the Following Problems 1. A 1.50 kg mass on a horizontal frictionless surface is attached to a horizontal spring with spring constant k = 200 N/m. The mass is in equilibrium at x = 0 m. The mass is released when t = 0 s at coordinate x = 0.100 m with a velocity (2.00 m/s)i ˆ. (a) Find the constants A, ω and φ in x(t) = Acos(ωt + φ) . (b) Find the period T of the oscillation. (c) Determine the maximum speed of the oscillation and the magnitude of the maximum acceleration. (d) Plot x(t) during the time interval t = 0 s and t = 2T. Solution: (a) From the problem, 133.3 N/kg m 1.5 200 = = = ⋅ m k ω , so substitute A, ω, t= 0 s, ,and x = 0.100 m and v=2.00 m/s to x(t) = Acos(ωt + φ) , we have ⎩ ⎨ ⎧ = = − = − = = ω φ φ φ 2 sin 133.3 sin 0.1 cos 0 0 v A A x A Solving the equations, we have ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = − = − = + = 3 3 5 ) 133.3 arctan( ) 0.1(m) 133.3 2 0.1 ( 0 0 2 2 π φ π or x v A Thinking of the direction of the velocity of the mass at t= 0 s, we get 3 π φ = − . N v f v F r mg v