Given block A = 50 kg and block B = 200 kg. The acceleration of gravity is 9.8 m/s2
If the pulleys have negligible mass and there is no friction, determine
a. The acceleration of blocks A and B
b. The tensions T1 and T2
Answers
Mass of block A = 50 kg
Mass of block B = 200 kg
Acceleration of block A = a₁ m/s²
Acceleration of block B = a₂ m/s²
Tension on string 1 attached to block A = T₁ N
Tension on string 2 attached to block B = T₂ N
From diagram ,
➳ T₂ = T₁ + T₁
➳ T₂ = 2T₁
As well ,
Work done by the Tension force is zero ,
➳ T₁(a₁) - T₂(a₂) = 0
➳ T₁.a₁ = T₂.a₂
➳ T₁.a₁ = (2T₁).a₂ [ ∵ T₂ = 2T₁ ]
➳ a₁ = 2a₂
Apply FBD for Block A :
➠ T₁ = 50a₁ ... (1)
Apply FBD for Block B :
➠ 200g - T₂ = 200a₂ ... (2)
➠ 200g - (2T₁) = 200a₂
➠ 200g - 2(50a₁) = 200a₂
➠ 200g - 100(2a₂) = 200a₂
➠ 400a₂ = 1960
➠ a₂ = 4.9 m/s²
Now ,
➠ a₁ = 2a₂
➠ a₁ = 2(4.9)
➠ a₁ = 9.8 m/s²
From (1) ,
➠ T₁ = 50(a₁)
➠ T₁ = 490 N
➠ T₂ = 2T₁
➠ T₂ = 2(490)
➠ T₂ = 980 N
Answer:
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Note : For seeing diagram need to see via web as picture environment is not working now in app
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Initial thrown velocity of the stone (u) = 29.4 m/s
Height of the tower (h) = 34.3 m
Total time taken to reach the foot of the tower (t) = ? s
Consider AB :
Initial velocity (u) = 29.4 m/s
Final velocity (v) = 0 m/s
Attains rest at maximum height
Acceleration (g) = - 9.8 m/s²
Against the gravity
★ Apply 1st equation of motion ,
➠ v = u + at
➠ 0 = 29.4 + (-9.8)t
➠ 9.8t = 29.4
➠ t = 3 s \pink{\star}⋆
★ Apply 3rd equation of motion ,
➠ v² - u² = 2as
➠ (0)² - (29.4)² = 2(-9.8)s
➠ - 864.36 = - 19.6s
➠ s = 44.1 m
Consider BD :
Initial velocity (u) = 0 m/s
Acceleration (g) = 9.8 m/s²
Starts from rest at maximum height
Total height , s' = h + s
➠ s' = 34.3 + 44.1
➠ s' = 78.4 m
Apply 2nd equation of motion ,
➠ s = ut + ¹/₂ at²
➠ 78.4 = (0)t + ¹/₂(9.8)t²
➠ 78.4 = 4.9t²
➠ t² = 16
➠ t = 4 s
Total time = 4 s + 3 s
➠ 7 s