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Sagot :
Answer:
 K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
    Em₀ = K = ½ mv²
final point. When it is at tea = 50º
    Em_f = K + U
    Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
    h = L - L cos 50
    Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
    Em₀ = Em_f
     ½ mv² = ½ m v_b² + mg L (1 - cos50)
     K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
     K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
     K_b = 36 +42.0
     K_b = 78 J
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