LAWS OF MOTION

Answer :

F = m × a

                   v² = u² + 2as

                   hence a = {(v² – u²) / 2s}

                   F = m {(v² – u²) / 2s}

                   F = (20) [{5² – 20²} / {2(100)}]

                   F = – 37.5 N   

Answer :

m= 0.2 kg.

                   Let velocity after band has released ball be v

                   v² = v_o² + 2as

                   v² = 2as

                   here v_o = 0, a = g, s = 2m

                   v² = (2)(10)(2) = 40

                   Force applied by hand causer ball to accelerate to this velocity from rest.

                             v² = v_o² + 2as

                             v² = 2as

                             a = (v² / 2s)

                             a = 40 / (2 × 0.2)

                             ∴ a = 100 m/s²

                   upward force     = F = m (g + a)

                                                = 0.2(10 + 100)

                                                = 22N  

Answer :

Newton's second law, Force = rate of change of linear momentum.

                             F = d(mv) / dt   

Answer :

P(t) = A [j˄ cos kt – j˄ sin kt]

                   F = [{dp(t)} / {dt}]

                   F = (d / dt) A [i˄ cos kt – j˄ sin kt]

                      = A [– i˄ sin kt (k)– j cos kt (k)]

                   F = Ak [– i˄ sin kt – j˄ cos kt]

                   angle between two vectors is θ

                   cos θ = {(F ∙ P) / (|F| |P|)}

                   = A [{(j˄ cos kt – j˄ sin kt) ∙ Ak (– i˄ sin kt – j˄ cos kt)} / {|F||P|}]

                   = A² k [{– sin kt cos kt + sin kt cos kt} / {|F| |P|}]

                   = A²k(0)

                   = 0

                   θ = cos^–1(0)

                   θ = 90°   

Answer :

F = 5N

                   W = 9.8 N

                   a = 2

                   w = mg = 9.8 N

                   m = {(9.8) / (9.8)} = 1 kg

                   F = ma

                   a = (F / m) = (5 / 1) = 5.00 m/s²