Iklan

Pertanyaan

Perhatikan katrol bergerak dibawah ini ! Apabila benda A bergerak keatas dan benda B bergerak ke bawah, maka persamaan percepatan yang benar adalah ….

Perhatikan katrol bergerak dibawah ini !

 

Apabila benda A bergerak keatas dan benda B bergerak ke bawah, maka persamaan percepatan yang benar adalah ….

  1. begin mathsize 14px style straight a subscript straight A equals 1 half straight a subscript straight B end style 

  2. begin mathsize 14px style straight a subscript straight A equals straight a subscript straight B end style 

  3. begin mathsize 14px style straight a subscript straight A equals 2 straight a subscript straight B end style 

  4. begin mathsize 14px style straight a subscript straight B equals 2 straight a subscript straight A end style 

  5. begin mathsize 14px style straight a subscript straight B equals 4 straight a subscript straight A end style 

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

00

:

13

:

50

:

49

Klaim

Iklan

A. Aulia

Master Teacher

Jawaban terverifikasi

Pembahasan

Diketahui : Ditanya : persamaan percepatan ? Jawab : panjang tali total saat kita meninjau kondisi awal yaitu : Sehingga : Kita buat permisalan : Persamaan di atas menjadi : ∆ y +2∆ s =0 Nah inget yah : Nah dari persamaan ini kita mendapatkan hasil : ∆ y =-2∆ s Nah tanda negatif (-) dari hasil ini hanya menunjukkan kalo ada perbedaan yang terjadi pada gerakan balok A dan balok B. Meninjau besarnya saja maka kita bisa tulis : ∆ y =2∆ s Menentukan perbandingan percepatan dari perbandian ketinggian kedua balok :

 

Diketahui :

begin mathsize 14px style OA equals straight y subscript 1 BC equals straight s subscript 1 DE equals straight s subscript 1 EF equals straight x DF equals straight s subscript 1 plus straight x straight O apostrophe straight A apostrophe equals straight y subscript 2 straight B apostrophe straight C apostrophe equals straight s subscript 2 straight D apostrophe straight E apostrophe equals straight s subscript 2 straight E apostrophe straight F apostrophe equals straight x straight D apostrophe straight F apostrophe equals straight s subscript 2 plus straight x end style 

Ditanya : persamaan percepatan ?

Jawab :

panjang tali total saat kita meninjau kondisi awal yaitu :

begin mathsize 14px style straight l equals OA plus AB with hat on top plus BC plus CD with hat on top plus DE plus EF straight l equals straight y subscript 1 plus AB with hat on top plus straight s subscript 1 plus CD with hat on top plus straight s subscript 1 plus straight x straight l equals straight y subscript 1 plus straight x plus 2 straight s subscript 1 plus AB with hat on top plus CD with hat on top Atau straight l equals straight O straight apostrophe straight A straight apostrophe plus stack straight A to the power of straight apostrophe straight B to the power of straight apostrophe with hat on top plus straight B straight apostrophe straight C straight apostrophe plus stack straight C to the power of straight apostrophe straight D to the power of straight apostrophe with hat on top plus straight D straight apostrophe straight E straight apostrophe plus straight E straight apostrophe straight F straight apostrophe straight l equals straight y subscript 2 plus stack straight A to the power of straight apostrophe straight B to the power of straight apostrophe with hat on top plus straight s subscript 2 plus stack straight C to the power of straight apostrophe straight D to the power of straight apostrophe with hat on top plus straight s subscript 2 plus straight x straight l equals straight y subscript 2 plus straight x plus 2 straight s subscript 2 plus stack straight A to the power of straight apostrophe straight B to the power of straight apostrophe with hat on top plus stack straight C to the power of straight apostrophe straight D to the power of straight apostrophe with hat on top end style 

Sehingga :

begin mathsize 14px style straight l equals straight l straight y subscript 1 plus straight x plus 2 straight s subscript 1 plus AB with hat on top plus CD with hat on top equals straight y subscript 2 plus straight x plus 2 straight s subscript 2 plus stack straight A to the power of straight apostrophe straight B to the power of straight apostrophe with hat on top plus stack straight C to the power of straight apostrophe straight D to the power of straight apostrophe with hat on top AB with hat on top equals stack straight A to the power of straight apostrophe straight B to the power of straight apostrophe with hat on top CD with hat on top equals stack straight C to the power of straight apostrophe straight D to the power of straight apostrophe with hat on top straight y subscript 1 plus straight x plus 2 straight s subscript 1 equals straight y subscript 2 plus straight x plus 2 straight s subscript 2 straight y subscript 1 plus 2 straight s subscript 1 equals straight y subscript 2 plus 2 straight s subscript 2 0 equals straight y subscript 2 plus 2 straight s subscript 2 minus straight y subscript 1 minus 2 straight s subscript 1 straight y subscript 2 minus straight y subscript 1 plus 2 straight s subscript 2 minus 2 straight s subscript 1 equals 0 straight y subscript 2 minus straight y subscript 1 plus 2 open parentheses straight s subscript 2 minus straight s subscript 1 close parentheses equals 0 end style 

Kita buat permisalan :

begin mathsize 14px style straight y subscript 2 minus straight y subscript 1 equals increment straight y straight s subscript 2 minus straight s subscript 1 equals increment straight s end style 

Persamaan di atas menjadi :

y+2∆s=0  

Nah inget yah :

begin mathsize 14px style increment straight y rightwards arrow berapa space jauh space balok space straight A space naik space ke space atas increment straight s rightwards arrow berapa space jauh space balok space straight B space turun space ke space bawah end style 

Nah dari persamaan ini kita mendapatkan hasil :

y=-2∆s  

Nah tanda negatif (-) dari hasil ini hanya menunjukkan kalo ada perbedaan yang terjadi pada gerakan balok A dan balok B.

Meninjau besarnya saja maka kita bisa tulis :

y=2∆s

Menentukan perbandingan percepatan dari perbandian ketinggian kedua balok :

begin mathsize 14px style straight h subscript straight A over straight h subscript straight B equals fraction numerator open parentheses 1 half straight a subscript straight A straight t squared close parentheses over denominator open parentheses 1 half straight a subscript straight B straight t squared close parentheses straight end fraction fraction numerator increment straight y over denominator increment straight s end fraction equals straight a subscript straight A over straight a subscript straight B fraction numerator 2 increment straight s over denominator increment straight s end fraction equals straight a subscript straight A over straight a subscript straight B straight a subscript straight A equals 2 straight a subscript straight B end style 

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

1

Iklan

Pertanyaan serupa

Perhatikan sistem di bawah ini. Jika dan m B ​ = 60 kg . Besar percepatan balok A adalah ...

3

0.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2025 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia