Sebuah benda bermassa m dilempatkan vertikal keatas selain mendapatkan gaya gravitasi mg benda terse.... Question from @nelviadesy

October 2018 1 37 Report

Sebuah benda bermassa m dilempatkan vertikal keatas selain mendapatkan gaya gravitasi mg benda tersebut mendapatkan gaya gesekan

mendokusai Gerak benda ke atas:

1. perlambatan:
Σ $F = m \times a_1$
$f + w = m \times a_1$
$\frac{1}{4} \times m \times g + m \times g = m \times a_1$
$\frac{5}{4} \times m \times g = m \times a_1$
$\frac{5}{4} \times g = a_1$
$a_1 = \frac{5}{4} \times g$

2. tinggi maksimum:
$v_1^2 = v_0^2 - 2 a_1 h$
$0^2 = v_0^2 - 2 a_1 h$
$0 = v_0^2 - 2 a_1 h$
$2 a_1 h = v_0^2$
$h = \frac{v_0^2}{2 a_1}$
$h = \frac{v_0^2}{2 \times \frac{5}{4} \times g}$
$h = \frac{v_0^2}{\frac{5}{2} \times g}$
$h = \frac{2 v_0^2}{5 g}$

gerak benda ke bawah:

1. percepatan:
Σ $F = m \times a_2$
$w - f = m \times a_2$
$m \times g - \frac{1}{4} \times m \times g = m \times a_2$
$\frac{3}{4} \times m \times g = m \times a_2$
$\frac{3}{4} \times g = a_2$
$a_2 = \frac{3}{4} \times g$

2. laju benda ketika mencapai permukaan tanah lagi:
$v_2^2 = v_1^2 + 2 a_2 h$
$v_2^2 = 0^2 + 2 \times \frac{3}{4} \times g \times \frac{2 v_0^2}{5 g}$
$v_2^2 = 2 \times \frac{3}{4} \times g \times \frac{2 v_0^2}{5 g}$
$v_2^2 = \frac{3}{5} v_0^2$
$v_2 = v_0 \sqrt{\frac{3}{5}}$
$v_2 = v_0 \sqrt{\frac{15}{25}}$
$v_2 = \frac{v_0}{5} \sqrt{15}$ m/s

Semoga membantu :)

19 votes Thanks 43

nelviadesy trm ksh ya..

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