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Pertanyaan

Aris bermain mobil mainan dengan lintasan seperti gambar berikut.

 

Mobil mainan mula-mula diletakkan di titik A dan bergerak dengan kecepatan awal 5 m/s melintas permukaan kasar AB (1 meter). Koefisien gesekan ban mobil mainan dengan lintasan AB adalah 0,3. Jika massa mobil mainan 50 gram dan R = 15 cm, tentukan kecepatan mobil mainan di titik C! begin mathsize 14px style left parenthesis g space equals space 10 space straight m divided by straight s squared right parenthesis end style 

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S. Syifa

Master Teacher

Jawaban terverifikasi

Jawaban

kecepatan benda di titik C adal 4,3 m/s

Pembahasan

Diketahui :

begin mathsize 14px style v subscript a equals 5 space straight m divided by straight s s equals 1 space straight m mu equals 0 comma 3 m equals 50 space gram equals 0 comma 05 space kg R equals 15 space cm equals 0 comma 15 space straight m end style  


Pembahasan :

Mencari kecepatan dititik B terlebih dahulu

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row W equals cell increment E k end cell row cell F cross times s end cell equals cell 1 half m left parenthesis v subscript b squared minus v subscript a squared right parenthesis end cell row cell negative mu cross times N cross times s end cell equals cell 1 half m left parenthesis v subscript b squared minus v subscript a squared right parenthesis end cell row cell negative 0 comma 3 cross times 0 comma 5 cross times 1 end cell equals cell 1 half cross times 0 comma 05 cross times left parenthesis v subscript b squared minus 5 squared right parenthesis end cell row cell negative 0 comma 15 end cell equals cell 0 comma 025 left parenthesis v subscript b squared minus 25 right parenthesis end cell row cell v subscript b squared minus 25 end cell equals cell negative 6 end cell row cell v subscript b squared end cell equals 19 row cell v subscript b end cell equals cell square root of 19 space straight m divided by straight s end cell row blank blank blank end table end style 

Menghitung kecepatan di titik C

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell E p subscript 0 plus E k subscript 0 end cell equals cell E p plus E k end cell row cell 0 plus 1 half m v subscript b squared end cell equals cell m g left parenthesis R minus R space cos space 37 right parenthesis plus 1 half m v subscript c squared end cell row cell 1 half cross times v b squared end cell equals cell g left parenthesis R minus R space cos space 37 right parenthesis plus 1 half v subscript c squared end cell row cell 1 half cross times 19 end cell equals cell 10 left parenthesis 0 comma 15 minus 0 comma 15 times 0 comma 8 right parenthesis plus 1 half v subscript c squared end cell row 19 equals cell 2 cross times 10 left parenthesis 0 comma 03 right parenthesis plus v subscript c squared end cell row cell v subscript c squared end cell equals cell 19 minus 0 comma 6 end cell row cell v subscript c squared end cell equals cell 18 comma 4 end cell row cell v subscript c end cell equals cell square root of 18 comma 4 end root end cell row blank equals cell 4 comma 3 space straight m divided by straight s end cell end table end style 

Jadi, kecepatan benda di titik C adal 4,3 m/s

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