Question

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Answer 1

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▷Hk. II Newton

• Anggap bidang licin.
m = 10 kg
tan α = ¾ ⇄ sin α = ⅗
g = 10 m/s²

a. Kasus I , bergerak dgn kec. tetap , F = ???

Berlaku :
∑ F = 0
F - m g sin α = 0
F = m g sin α
F = 10 × 10 × ⅗
F = 60 N

b. Kasus II , benda bergerak ke atas dgn a = 2 m/s²

Berlaku :
∑ F = m a
F - m g sin α = m a
F = m (a + g sin α)
F = 10 (2 + 10 (⅗) )
F = 80 N

c. Kasus III , benda bergerak ke bawah dgn a = 2 m/s²

Berlaku :
∑ F = m a
F - m g sin α = m (-a)
F = m (g sin α - a)
F = 10 (10 (⅗) - 2)
F = 40 N