Roboguru

Dua buah benda A dan B masing-masing bermassa , jatuh dari ketinggian dan . Jika A menyentuh tanah dengan kecepatan , benda B akan menyentuh tanah dengan energi kinetik sebesar....

Pertanyaan

Dua buah benda A dan B masing-masing bermassa begin mathsize 14px style m end style, jatuh dari ketinggian begin mathsize 14px style h space meter end style dan begin mathsize 14px style 4 h space meter end style. Jika A menyentuh tanah dengan kecepatan begin mathsize 14px style v space straight m divided by straight s end style, benda B akan menyentuh tanah dengan energi kinetik sebesar....

  1. begin mathsize 14px style 4 m v squared end style 

  2. begin mathsize 14px style 2 m v squared end style 

  3. begin mathsize 14px style 3 over 4 m v squared end style 

  4. begin mathsize 14px style 1 half m v squared end style 

  5. begin mathsize 14px style 1 fourth m v squared end style 

Pembahasan Video:

Pembahasan Soal:

Diketahui :
begin mathsize 14px style m subscript A equals m subscript B equals m h subscript A equals space h space meter h subscript B equals 4 h space meter v subscript A equals straight v space straight m divided by straight s end style

Ditanya:
begin mathsize 14px style E k subscript B equals... ? end style

Penyelesaian :

KIta gunakan persamaan dari Hukum Kekekalan Energi Mekanik dimana benda awalnya dijatuhkan dari ketinggian:

begin mathsize 14px style thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space E M subscript 1 equals E M subscript 2 E p subscript 1 plus E k subscript 1 equals E p subscript 2 plus E k subscript 2 thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space E p subscript 1 equals E k subscript 2 end style

Sehingga, dapat kita gunakan perbandingan dari kedua benda:

begin mathsize 14px style thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space fraction numerator E p subscript 1 A end subscript over denominator E p subscript 1 B end subscript end fraction equals fraction numerator E k subscript 2 A end subscript over denominator E k subscript 2 B end subscript end fraction fraction numerator m subscript A cross times g cross times h subscript A over denominator m subscript B cross times g cross times h subscript B end fraction equals fraction numerator begin display style bevelled fraction numerator 1 over denominator 2 cross times m subscript A cross times v subscript A squared end fraction end style over denominator E k subscript B end fraction thin space thin space thin space fraction numerator up diagonal strike m cross times up diagonal strike g cross times up diagonal strike h over denominator up diagonal strike m cross times up diagonal strike g cross times 4 up diagonal strike h end fraction equals fraction numerator begin display style bevelled fraction numerator 1 over denominator 2 cross times m cross times v squared end fraction end style over denominator E k subscript B end fraction thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space 1 fourth equals fraction numerator begin display style bevelled fraction numerator 1 over denominator 2 cross times m cross times v squared end fraction end style over denominator E k subscript B end fraction thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space E k subscript B equals 4 cross times 1 half cross times m cross times v squared thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space E k subscript B equals 2 m v squared end style


Jadi, jawaban yang benar adalah (B).

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 April 2021

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