פתרון בגרות מכניקה — שאלון 36361

נקודה | A | B | C | D
t (s) | 0 | 1 | 3 | 4
y (m) | 40 | 45 | 25 | 0
g = 10 m/s² ; y₀ = 40 m

0.00 s

0.00 s

40.00 m

10.00 m/s

10.00 m/s²

כלפי מעלה

0.00 m

10.00 m/s

y(t) = 40 + 10·t − 5·t² m v(t) = 10 − 10·t m/s a = − 10 m/s²

a = − g = − 10 m/s² y(t) = y₀ + v₀·t − 12g·t² v(t) = v₀ − g·t

v(t_B) = v₀ − g·t_B = 0 v₀ = g·t_B = 10 · 1 = 10 m/s

45 = 40 + v₀ · 1 − 12 · 10 · 1² = 35 + v₀ v₀ = 45 − 35 = 10 m/s

0 = 40 + v₀ · 4 − 12 · 10 · 4² = 40 + 4·v₀ − 80 4·v₀ = 40 , v₀ = 10 m/s y(t) = 40 + 10·t − 5·t²

v_A = 10 − 10 · 0 = + 10 m/s v_C = 10 − 10 · 3 = − 20 m/s

v_C² = v₀² − 2·g·(y_C − y₀) = 100 − 2 · 10 · (25 − 40) = 400 |v_C| = √400 = 20 m/s

a = Σ Fm = m·gm = g = 10 m/s²

12a = − 5 , a = − 10 m/s²

v_ave = y_D − y_At_D − t_A = 0 − 404 − 0 = − 10 m/s

Δy = + 5 − 45 = − 40 m , v_ave = − 404 = − 10 m/s

v_ave = v₀ + v_D2 = 10 + (− 30)2 = − 10 m/s

504 = 12.5 m/s

v(t) = v₀ − g·t = 10 − 10·t m/s slope = ΔvΔt = − 30 − (+ 10)4 − 0 = − 10 m/s²

a (0 , + 10) ; b (1 , 0) ; c (3 , − 20) ; d (4 , − 30)

|v_D| = √v₀² + 2·g·y₀ = √100 + 800 = 30 m/s |v_C| = √100 + 2 · 10 · 15 = √400 = 20 m/s

y_C = 25 m ; v_y,C = − 20 m/s ; a_y = − 10 m/s²

t (s) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6
v (m/s) | 0.45 | 0.70 | 1.15 | 1.50 | 1.95 | 2.25
m₁ = 0.5 kg ; α = 30° ; g = 10 m/s² ; μ = 0

0.00 s

0.00 s

0.00 m/s

3.75 m/s²

1.00

4.375 N

4.33 N

0.000 m

T − m₁·g·sin α = m₁·a m₂·g − T = m₂·a N = m₁·g·cos α = 4.33 N

(0.1 , 0.45) (0.2 , 0.70) (0.3 , 1.15) (0.4 , 1.50) (0.5 , 1.95) (0.6 , 2.25)

slope = ΔvΔt = 2.25 − 0.450.6 − 0.1 = 1.800.50 = 3.6 m/s²

slope = Σ (t_i − t̄)(v_i − v̄)Σ (t_i − t̄)² = 0.6550.175 = 3.743 m/s² v₀ = 1.3333 − 3.743 · 0.35 = 0.023 m/s

slope = 0.360.1 = 3.6 m/s²

T − m₁·g·sin 30° = m₁·a N − m₁·g·cos 30° = 0 N = 0.5 · 10 · √32 = 4.33 N

m₂·g − T = m₂·a

T = m₁·(a + g·sin 30°) = 0.5 · (3.75 + 5) = 0.5 · 8.75 = 4.375 N

m₂ = m₁·(a + g·sin 30°)g − a = 4.37510 − 3.75 = 4.3756.25 = 0.70 kg

a = (m₂ − m₁·sin 30°)·gm₁ + m₂ = 0.45 · 101.2 = 4.51.2 = 3.75 m/s²

− m₁·g·sin 30° = m₁·a₁ , a₁ = − g·sin 30° = − 5 m/s²

a₂ = g = + 10 m/s² a₁a₂ = − g·sin 30°g = − sin 30° = − 510 = − 0.5

a₁ = 2.50.5 = 5 m/s² ; a₂ = 70.7 = 10 m/s²

Δt = 3.755 = 0.75 s ; Δs = 3.75²2 · 5 = 1.40625 m

|a₂||a₁| = 105 = 2

r = 3 m ; μ_s = 0.6 ; g = 10 m/s²
m₁ = 70 kg ; m₂ = 90 kg ; ω = 2.6 1/s
f = m·g ; N = m·ω²·r ; f ≤ μ_s·N

2.600 1/s

90 kg

0.60

3.0 m

2.357 1/s

1825.20 N

1095.12 N

900.00 N

82.2 %

20.28 m/s²

7.80 m/s

2.417 s

0.4931

m·g = 70 · 10 = 700 N

Σ F_y = f − m·g = m·a_y = 0 f = m·g

a_c = ω²·r = v²r Σ F_x = N = m·a_c = m·ω²·r = m·v²r

f ≤ μ_s·N

m·g ≤ μ_s · m·ω²·r g ≤ μ_s·ω²·r ω² ≥ gμ_s·r

ω_min = √gμ_s·r = √100.6 · 3 = √101.8 ω_min = √5.5556 = 2.357 1/s

N = m·v²r ; m·g = μ_s · m·v²r v_min = √g·rμ_s = √10 · 30.6 = √50 = 7.071 m/s ω_min = v_minr = 7.0713 = 2.357 1/s

T = 2πω_min = 2.666 s ; 1T = 0.375 1/s a_c = ω_min²·r = 5.5556 · 3 = 16.67 m/s² ≈ 1.67·g

N = 70 · 5.5556 · 3 = 1166.67 N μ_s·N = 0.6 · 1166.67 = 700 N = m·g = 700 N

ω_min = √gμ_s·r

m·g = 90 · 10 = 900 N μ_s·N = μ_s · m · ω_min² · r = 0.6 · 90 · 5.5556 · 3 = 900 N

f − m·g = 0 ⟹ f = m·g = 90 · 10 = 900 N

N = m·ω²·r = 90 · 6.76 · 3 = 1825.2 N f_max = μ_s·N = 0.6 · 1825.2 = 1095.12 N

μ = fN = 9001825.2 = 0.4931

ff_max = gμ_s·ω²·r = ω_min²ω² = 0.8218 f = 0.8218 · 1095.12 = 900 N

m = 2 kg ; h = 0.8 m ; BD = 4 m ; g = 10 m/s²
x (m) | 0 | 1 | 2 | 3 | 4
F (N) | 0 | 2 | 4 | 6 | 8
k = 2 N/m ; F(x) = 2·x ; v₀ = 5 m/s

5.0 m/s

0.00 m

5.00 m/s

0.00 N

0.00 m/s²

25.00 J

0.00 J

0.00 s

4.00 m/s

m·v_A²2 + 0 = m·v_B²2 + m·g·h ; v_B = 0

v_A = √2·g·h = √2 · 10 · 0.8 = √16 = 4 m/s

v_B² = v_A² − 2·g·sinθ·L ; L = hsinθ ⟹ v_B² = v_A² − 2·g·h

m·g·h = 2 · 10 · 0.8 = 16 J m·v_A²2 = 2 · 4²2 = 16 J

E_k(A) = m·v_0²2 = 2 · 5²2 = 25 J

E_k(B) = 25 − m·g·h = 25 − 16 = 9 J v_B = √2 · E_k(B)m = √9 = 3 m/s

W_BC = E_k(C) − E_k(B) = 0 − 9 = −9 J

E_k(C) − E_k(A) = W_g + W_F 0 − 25 = −16 + W_F ⟹ W_F = −9 J

k = ΔFΔx = 8 − 04 − 0 = 2 N/m ⟹ F(x) = 2·x

|W| = d · F(d)2 = d · 2·d2 = d²

d² = 9 ⟹ d = √9 = 3 m

W = −∫ 2·x dx = −[x²] = −d² = −9 J ⟹ d = 3 m

|W| = F_avg · d = 3 · 3 = 9 J

a(x) = F(x)m = 2·x2 = x a(C) = 62 = 3 m/s² ; a(B) = 02 = 0

W = 9 − x² = m·v²2 ⟹ v(x) = √9 − x²

ω = √km = √22 = 1 1/s x(t) = 3·sin(t) ; v(t) = 3·cos(t) ; t_C = π2 = 1.571 s

W(B..C) + W(C..B) = −9 + 9 = 0 J

m·v_A²2 = 25 ⟹ v_A² = 25 ⟹ v_A = 5 m/s

v_A = √v_B² + 2·g·h = √3² + 2 · 10 · 0.8 = √25 = 5 m/s

T = 2π√m_A + m_Bk | m_A = 1 kg | (1 , 0) (2 , 3) (3 , 8)
(1) F_sp = 1 (2) F_sp = 2 (3) F_sp = 1 (4) F_sp = 4 m·g = 2

המסה מעל נקודת שיווי המשקל · הכוח השקול כלפי מטה

3.0 kg

1.50

39.4784

4.00

2.0000

3.1416

0.9929

1.4894

1.500

−19.60

−58.80

−14.70

10.00

1.00

0.0200

T = 2π√m_A + m_Bk m_B = k·T²4π² − m_A k = 4π²·(m_A + m_B)T²

שתי משבצות שקולות ל-mg · הסימולציה נפתחת בהעתק של תרשים (1); לחצו על שחרר כדי להתחיל את התנודה

t₁ = T ΔtT = 0.22.0 = 0.10 = 10 %

t₁₀ = 10·T Δt10·T = 0.220.0 = 0.01 = 1 %

T = t₁₀10 ΔT = Δt10 = 0.02 s

ΔTT = Δt₁₀t₁₀ = Δt10·T

m = m_A + m_B T = 2π√m_A + m_Bk

T² = 4π²·(m_A + m_B)k m_A + m_B = k·T²4π²

m_B = k·T²4π² − m_A

m_A + m_B = m_A + k·T²4π² − m_A = k·T²4π² 2π√k·T²4π²·k = 2π·T2π = T

[k·T²] = kgs² · s² = kg

T_min = 2π√m_Ak

T_min = 1.0 s

T = 2π√m_A + m_Bk ≥ 2π√m_Ak = 1.0 s

m_B = k·T²4π² − m_A = T² − 1 < 0

k = 4π²·(m_A + m_B)T²

k = 4π²·(1 + 0)1.0² = 4π² = 39.4784 ≈ 39.5 N/m

k = 4π²·(1 + 3)2.0² = 4π²·44 = 4π² = 39.5 N/m

k = 4π²·(1 + 8)3.0² = 4π²·99 = 4π² = 39.5 N/m

m_B = k4π²·T² − m_A slope = 3.0 − 04.0 − 1.0 = 1.0 kg/s² k = 4π²·1.0 = 39.4784 ≈ 39.5 N/m

m_B = T² − 1

ΣF = F_sp − m·g = −k·x

ΣF = 0.5·m·g + m·g = 1.5·m·g x = 1.5·m·gk ; a = 1.5·g

F_sp = m·g ; ΣF = 0 ; a = 0 ; x = 0

F_sp = 0.5·m·g ; ΣF = 0.5·m·g x = 0.5·m·gk ; a = 0.5·g

F_sp = 2·m·g ; ΣF = m·g x = − m·gk ; a = g

A ≥ 1.5·m·gk

G = 6.67·10⁻¹¹ N·m²/kg² | M_E = 5.97·10²⁴ kg | R_E = 6.37·10⁶ m
M_M = 7.35·10²² kg | R_M = 1.74·10⁶ m | d = 3.84·10⁸ m | h = 1.9·10⁵ m

מסלול הביניים · הכוח השקול הוא כוח הכבידה, והוא רדיאלי אל מרכז כדור הארץ

190 km

15000 kg

0.400

81.22

6.560·10⁶

7791.09

7.791

5290.4

88.2

9.2532

9.8134

11018.26

4.553·10¹¹

−9.105·10¹¹

−4.553·10¹¹

1.536·10⁸

253.168

1.385

251.783

3.456·10⁸

90.01

9.012

v = √G·M_Er E = − G·M_E·m2·r ΣF_G = G·m·M_Er_E² − G·m·M_Mr_M²

לחצו על אחד מארבעת הגרפים · הם ממוספרים מימין לשמאל, כמו בשאלון

r = R_E + h = 6.37·10⁶ + 1.9·10⁵ = 6.56·10⁶ m

G·M_E·mr² = m·v²r v = √G·M_Er

G·M_E = 6.67·10⁻¹¹ · 5.97·10²⁴ = 3.98199·10¹⁴ v² = 3.98199·10¹⁴6.56·10⁶ = 6.0701067·10⁷ v = 7791.09 ≈ 7.79·10³ m/s

T = 2π·rv = 5290.4 s = 88.2 min

v²r = G·M_Er² = 9.2532 m/s² g₀ = 9.8134 m/s² ; R_E²r² = 0.9429

a = v²r = 9.2532 m/s² ΣF = G·M_E·mr² ≠ 0

v = √G·M_Er = 7791 m/s

v = √a·r = √9.2532 · 6.56·10⁶ = 7791 m/s

E_k = m·v²2 = + G·M_E·m2·r E_p = − G·M_E·mr E = E_k + E_p = − G·M_E·m2·r

E = −3.0350533·10⁷ · m J E (m = 1000) = −3.035·10¹⁰ J ; E (m = 2000) = −6.070·10¹⁰ J

v_esc = √2 · v = 11018.26 m/s

F_E = G·M_E·mr_E² ; F_M = G·M_M·mr_M²

ΣF_G = G·m·M_Er_E² − G·m·M_Mr_M² ΣF_G = G·m·( M_Er_E² − M_Mr_M² )

M_Er_E² = M_Mr_M² r_Er_M = √M_EM_M = √81.22 = 9.012 r_E = 0.9001 · d = 3.456·10⁸ m

פתרון בגרות · מכניקה