Jawaban:

Penjelasan:

Diket:

m = 20 kg

F = 1200 N

α = 30°

β = 60°

Ditanya:

A. Percepatan benda saat menaiki bidang miring

B. Percepatan benda jika bidang miring kasar

Dijawab:

A. ∑Fx = m.a

F cos β - w sin α = m . a

a = F cos β - w sin α / m

a = 1200 1/2 - 200 1/2 / 20

a = 600 - 100 /20

a = 500/20

a = 25 m/s²

B. ∑Fy = 0

N - w cos α - F sin β = 0

N = 200 . 1/2 √3 + 1200 1/2 √3

N = 100 √3 + 600 √3

N = 700 √3

∑Fx= m.a

F cos β - w sin α - fg = m.a

a = F cos β - w sin α - fg / m

a = 1200 1/2 - 200 1/2 - 700√3 . √3/25 / 20

a = 600 - 100 - 84  / 20

a = 20,8 m/s²