{"id":310690,"date":"2023-01-11T09:05:42","date_gmt":"2023-01-11T02:05:42","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=310690"},"modified":"2023-05-09T15:12:06","modified_gmt":"2023-05-09T08:12:06","slug":"contoh-soal-hukum-newton-2","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/mapel\/fisika\/contoh-soal-hukum-newton-2\/","title":{"rendered":"Kumpulan Contoh Soal Hukum Newton 2 Lengkap dengan Jawaban dan Penjelasan"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

Hai Quipperian, di artikel sebelumnya Quipper Blog sudah pernah membahas tentang Hukum Newton 2 beserta penerapannya. Apa sih<\/em> yang kamu ingat dari hukum ini? Ya benar. Hukum Newton 2 merupakan hukum yang membahas hubungan antara gaya dan percepatan gerak benda. Agar kamu tidak lupa, yuk simak contoh soal Hukum Newton 2 berikut ini!<\/p>\n\n\n\n

Contoh Soal 1<\/strong><\/h2>\n\n\n\n

Sebuah bola bermassa 450 gram dipukul dengan gaya 2 N. Akibatnya benda yang semula diam menjadi bergerak. Percepatan gerak benda tersebut adalah \u2026.<\/p>\n\n\n\n

    \n
  1. 4,4 m\/s2<\/sup><\/li>\n\n\n\n
  2. 1,3 m\/s2<\/sup><\/li>\n\n\n\n
  3. 2,7 m\/s2<\/sup><\/li>\n\n\n\n
  4. 3,8 m\/s2<\/sup><\/li>\n\n\n\n
  5. 5,4 m\/s2<\/sup><\/li>\n<\/ol>\n\n\n\n

    Pembahasan:<\/p>\n\n\n\n

    Diketahui:<\/p>\n\n\n\n

    m<\/em> = 450 gr = 0,45 kg<\/p>\n\n\n\n

    F<\/em> = 2 N<\/p>\n\n\n\n

    Ditanya: a<\/em> =\u2026?<\/p>\n\n\n\n

    Jawab:<\/p>\n\n\n\n

    Untuk mencari percapatan gerak bola, gunakan persamaan Hukum Newton 2 berikut ini.<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Jadi, percepatan gerak bola tersebut adalah 4,44 m\/s2<\/sup>.<\/p>\n\n\n\n

    Jawaban: A<\/p>\n\n\n\n

    Contoh Soal 2<\/strong><\/h2>\n\n\n\n

    Sebuah balok kayu bermassa 5 kg diam di atas lantai yang licin. Lalu, balok kayu tersebut diberi gaya mendatar 10 N hingga bergerak. Besarnya kecepatan balok kayu setelah bergerak selama 2 s adalah \u2026.<\/p>\n\n\n\n

      \n
    1. 5 m\/s<\/li>\n\n\n\n
    2. 6 m\/s<\/li>\n\n\n\n
    3. 4 m\/s<\/li>\n\n\n\n
    4. 2,5 m\/s<\/li>\n\n\n\n
    5. 3 m\/s<\/li>\n<\/ol>\n\n\n\n

      Pembahasan:<\/p>\n\n\n\n

      Diketahui:<\/p>\n\n\n\n

      m<\/em> = 5 kg<\/p>\n\n\n\n

      F<\/em> = 10 N<\/p>\n\n\n\n

      t <\/em>= 2 s<\/p>\n\n\n\n

      v<\/em>0 <\/sub>= 0 (diam)<\/p>\n\n\n\n

      Ditanya: v<\/em>t<\/sub> saat t<\/em> = 2 s =\u2026?<\/p>\n\n\n\n

      Jawab:<\/p>\n\n\n\n

      Mula-mula, kamu harus mencari percepatan gerak balok kayu tersebut setelah dikenai gaya 10 N.<\/p>\n\n\n\n

      \"\"<\/figure>\n\n\n\n

      Selanjutnya, substitusikan nilai a<\/em> = 2 m\/s2<\/sup> ke persamaan kecepatan GLBB seperti berikut.<\/p>\n\n\n\n

      \"\"<\/figure>\n\n\n\n

      Jadi, kecepatan balok kayu setelah bergerak selama 2 s adalah 4 m\/s.<\/p>\n\n\n\n

      Jawaban: C<\/p>\n\n\n\n

      Contoh Soal 3<\/strong><\/h2>\n\n\n\n

      Sebuah benda ditarik dengan tiga gaya yang arahnya seperti gambar berikut.<\/p>\n\n\n\n

      \"\"\/<\/figure>\n\n\n\n

      Akibat adanya gaya tarikan di kedua sisinya, benda bergerak dengan percepatan 0,5 m\/s2<\/sup>. Massa benda tersebut adalah \u2026.<\/p>\n\n\n\n

        \n
      1. 10 kg<\/li>\n\n\n\n
      2. 8 kg<\/li>\n\n\n\n
      3. 4 kg<\/li>\n\n\n\n
      4. 16 kg<\/li>\n\n\n\n
      5. 12 kg<\/li>\n<\/ol>\n\n\n\n

        Pembahasan:<\/p>\n\n\n\n

        Mula-mula, kamu harus mencari resultan gaya yang bekerja pada benda. Ingat, gaya merupakan besaran vektor, sehingga arah harus diperhatikan.<\/p>\n\n\n\n

        Besar resultan gayanya dirumuskan sebagai berikut.<\/p>\n\n\n\n

        \"\"<\/figure>\n\n\n\n

        Lalu, substitusikan besar resultan gaya tersebut pada persamaan Hukum Newton 2.<\/p>\n\n\n\n

        \"\"<\/figure>\n\n\n\n

        Jadi, massa benda tersebut adalah 16 kg.<\/p>\n\n\n\n

        Jawaban: D<\/p>\n\n\n\n

        \n

        Baca Juga: <\/span> Rumus Resultan Gaya Searah, Berlawanan, Tegak Lurus dan Contoh Soal<\/a><\/p>\n<\/div>\n\n\n\n\n

        Contoh Soal 4<\/strong><\/h2>\n\n\n\n

        Pak Rosid memiliki dua buah balok, yaitu balok x dan balok y. Kedua balok itu diletakkan di atas lantai licin dan dihubungkan dengan tali seperti gambar berikut.<\/p>\n\n\n\n

        \"\"\/<\/figure>\n\n\n\n

        Untuk memindahkan kedua balok tersebut, Pak Rosid menariknya dengan gaya 20 N. Jika massa balok x dan y berturut-turut adalah 2 kg dan 8 kg, gaya tegangan tali antara kedua balok saat ditarik adalah \u2026.<\/p>\n\n\n\n

          \n
        1. 4 N<\/li>\n\n\n\n
        2. 5 N<\/li>\n\n\n\n
        3. 6 N<\/li>\n\n\n\n
        4. 8 N<\/li>\n\n\n\n
        5. 4,5 N<\/li>\n<\/ol>\n\n\n\n

          Pembahasan:<\/p>\n\n\n\n

          Diketahui:<\/p>\n\n\n\n

          F<\/em> = 20 N<\/p>\n\n\n\n

          m<\/em>x <\/sub>= 2 kg<\/p>\n\n\n\n

          m<\/em>y<\/sub> = 8 kg<\/p>\n\n\n\n

          Ditanya: T<\/em> =\u2026?<\/p>\n\n\n\n

          Jawab:<\/p>\n\n\n\n

          Mula-mula, uraikan komponen gaya yang bekerja pada balok seperti berikut.<\/p>\n\n\n\n

          \"\"\/<\/figure>\n\n\n\n

          Lalu, tentukan percepatan sistemnya.<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          Selanjutnya, tentukan tegangan talinya dengan meninjau balok x.<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          Jika kamu ingin meninjau dari balok y juga bisa, caranya seperti berikut.<\/p>\n\n\n\n

          \"\"<\/figure>\n\n\n\n

          Jadi, gaya tegangan tali antara kedua balok saat ditarik adalah 4 N.<\/p>\n\n\n\n

          Jawaban: A<\/p>\n\n\n\n

          Contoh Soal 5<\/strong><\/h2>\n\n\n\n

          Seseorang yang massanya 65 kg berada di dalam lift yang bergerak ke bawah dengan percepatan 1,5 m\/s2<\/sup>. Besarnya gaya normal yang dikerjakan lantai lift pada kaki seseorang tersebut adalah \u2026.<\/p>\n\n\n\n

            \n
          1. 425,5 N<\/li>\n\n\n\n
          2. 552,5 N<\/li>\n\n\n\n
          3. 320,5 N<\/li>\n\n\n\n
          4. 500,4 N<\/li>\n\n\n\n
          5. 338,5 N<\/li>\n<\/ol>\n\n\n\n

            Pembahasan:<\/p>\n\n\n\n

            Diketahui:<\/p>\n\n\n\n

            m<\/em> = 65 kg<\/p>\n\n\n\n

            a<\/em> = 1,5 m\/s2<\/sup><\/p>\n\n\n\n

            g<\/em> = 10 m\/s2<\/sup><\/p>\n\n\n\n

            Ditanya: N<\/em> =\u2026?<\/p>\n\n\n\n

            Jawab:<\/p>\n\n\n\n

            Untuk memudahkanmu dalam mengerjakan soal tersebut, uraikan dahulu komponen gaya yang bekerja pada seseorang tersebut.<\/p>\n\n\n\n

            \"\"\/<\/figure>\n\n\n\n

            Gaya yang searah dengan gerak lift diberi tanda positif dan sebaliknya, lalu gunakan persamaan Hukum Newton 2. Dengan demikian:<\/p>\n\n\n\n

            \"\"<\/figure>\n\n\n\n

            Jadi, besarnya gaya normal yang dikerjakan lantai lift pada kaki orang tersebut adalah 552,5 N.<\/p>\n\n\n\n

            Jawaban: B<\/p>\n\n\n\n

            Contoh Soal 6<\/strong><\/h2>\n\n\n\n

            Pak Hendra yang massanya 70 kg mengendarai mobil bermassa 1.030 kg dengan kecepatan 20 m\/s. Saat asyik berkendara, tiba-tiba ada sebuah HP yang terjatuh tepat 4 m di depan mobil Pak Hendra. Hal itu sontak membuat Pak Hendra kaget hingga mengerem mobilnya. Besarnya gaya pengereman minimal agar mobil Pak Hendra tidak menabrak HP tersebut adalah \u2026.<\/p>\n\n\n\n

              \n
            1. 27 kN<\/li>\n\n\n\n
            2. 32 kN<\/li>\n\n\n\n
            3. 55 kN<\/li>\n\n\n\n
            4. 72 kN<\/li>\n\n\n\n
            5. 80 kN<\/li>\n<\/ol>\n\n\n\n

              Pembahasan:<\/p>\n\n\n\n

              Diketahui:<\/p>\n\n\n\n

              v<\/em>0<\/sub><\/em> <\/em> = 20 m\/s<\/p>\n\n\n\n

              s<\/em> = 4 m<\/p>\n\n\n\n

              v<\/em>t<\/sub><\/em> = 0 (berhenti)<\/p>\n\n\n\n

              m<\/em>H<\/sub> = 70 kg<\/p>\n\n\n\n

              m<\/em>M<\/sub><\/em> = 1.030 kg<\/p>\n\n\n\n

              Ditanya: F<\/em> =\u2026?<\/p>\n\n\n\n

              Jawab:<\/p>\n\n\n\n

              Syarat agar HP tidak tertabrak adalah mobil Pak Hendra harus berhenti tepat di depan HP tersebut. Mula-mula, tentukan dahulu percepatan mobil Pak Hendra dengan rumus GLBB berikut.<\/p>\n\n\n\n

              \"\"<\/figure>\n\n\n\n

              Tanda negatif artinya terjadi perlambatan akibat pengereman.<\/p>\n\n\n\n

              Selanjutnya, substitusikan nilai percepatan di atas pada persamaan Hukum Newton 2.<\/p>\n\n\n\n

              \"\"<\/figure>\n\n\n\n

              Jadi, besarnya gaya pengereman minimal agar mobil Pak Hendra tidak menabrak HP tersebut adalah 55 kN.<\/p>\n\n\n\n

              Jawaban: C<\/p>\n\n\n\n

              Contoh Soal 7<\/strong><\/h2>\n\n\n\n

              Sebuah benda didorong dengan gaya 10 N di lantai datar yang kasar dengan kecepatan 5 m\/s. Jika koefisien gesek kinetis lantai tersebut adalah 0,2 dan massa bendanya 3 kg, benda akan berhenti setelah bergerak selama \u2026.<\/p>\n\n\n\n

                \n
              1. 5 s<\/li>\n\n\n\n
              2. 10 s<\/li>\n\n\n\n
              3. 15 s<\/li>\n\n\n\n
              4. 20 s<\/li>\n\n\n\n
              5. 12 s<\/li>\n<\/ol>\n\n\n\n

                Pembahasan:<\/p>\n\n\n\n

                Diketahui:<\/p>\n\n\n\n

                v<\/em>0<\/sub><\/em> = 5 m\/s<\/p>\n\n\n\n

                \u03bck<\/sub> = 0,2<\/p>\n\n\n\n

                m<\/em> = 3 kg<\/p>\n\n\n\n

                F<\/em> = 10 N<\/p>\n\n\n\n

                Ditanya: t<\/em> =\u2026?<\/p>\n\n\n\n

                Jawab:<\/p>\n\n\n\n

                Mula-mula, kamu harus mencari percepatan gerak benda menggunakan persamaan Hukum Newton 2.<\/p>\n\n\n\n

                \"\"<\/figure>\n\n\n\n

                Lalu, substitusikan nilai percepatan di atas pada persamaan GLBB diperlambat.<\/p>\n\n\n\n

                \"\"<\/figure>\n\n\n\n

                Jadi, benda akan berhenti setelah bergerak selama 15 s.<\/p>\n\n\n\n

                Jawaban: C<\/p>\n\n\n\n

                Contoh Soal 8<\/strong><\/h2>\n\n\n\n

                Sebuah balok bermassa 4 kg bergerak pada papan bidang miring yang kasar seperti berikut.<\/p>\n\n\n\n

                \"\"\/<\/figure>\n\n\n\n

                Sudut yang dibentuk antara papan bidang miring dan sisi mendatarnya adalah 30o<\/sup>. Jika koefisien gesek kinetik antara papan dan balok adalah 0,23, percepatan gerak balok tersebut adalah \u2026.<\/p>\n\n\n\n

                  \n
                1. 3,2 m\/s2<\/sup><\/li>\n\n\n\n
                2. 1,6 m\/s2<\/sup><\/li>\n\n\n\n
                3. 2,5 m\/s2<\/sup><\/li>\n\n\n\n
                4. 2 m\/s2<\/sup><\/li>\n\n\n\n
                5. 3 m\/s2<\/sup><\/li>\n<\/ol>\n\n\n\n

                  Pembahasan:<\/p>\n\n\n\n

                  Diketahui:<\/p>\n\n\n\n

                  m<\/em> = 4 kg<\/p>\n\n\n\n

                  \u03bck<\/sub> = 0,23<\/p>\n\n\n\n

                  \u03b8 = 30o<\/sup><\/p>\n\n\n\n

                  g<\/em> = 10 m\/s2<\/sup><\/p>\n\n\n\n

                  Ditanya: a<\/em> =\u2026?<\/p>\n\n\n\n

                  Jawab:<\/p>\n\n\n\n

                  Mula-mula, uraikan komponen gaya yang bekerja pada sistem bidang miring tersebut.<\/p>\n\n\n\n

                  \"\"\/<\/figure>\n\n\n\n

                  Oleh karena tidak ada gerak benda pada sumbu-y, maka <\/sub>. Dengan demikian:<\/p>\n\n\n\n

                  \"\"<\/figure>\n\n\n\n

                  Dari gambar di atas terlihat bahwa benda bergerak pada komponen sumbu-x. Artinya, pada sumbu-x berlaku persamaan Hukum Newton 2.<\/p>\n\n\n\n

                  \"\"<\/figure>\n\n\n\n

                  Selanjutnya, substitusikan nilai N<\/em> pada persamaan (1) ke persamaan (2).<\/p>\n\n\n\n

                  \"\"<\/figure>\n\n\n\n

                  Jadi, percepatan gerak balok tersebut adalah 2 m\/s2<\/sup>.<\/p>\n\n\n\n

                  Jawaban: D<\/p>\n\n\n\n

                  Contoh Soal 9<\/strong><\/h2>\n\n\n\n

                  Sebuah mobil-mobilan bermassa 500 gr diam di atas lantai yang licin. Lalu, adik mendorong mobil-mobilannya tersebut dengan gaya 0,25 N. Jarak yang ditempuh mobil-mobilan setelah adik mendorongnya selama 3 s adalah \u2026.<\/p>\n\n\n\n

                    \n
                  1. 2,25 m<\/li>\n\n\n\n
                  2. 3,25 m<\/li>\n\n\n\n
                  3. 1,75 m<\/li>\n\n\n\n
                  4. 2,6 m<\/li>\n\n\n\n
                  5. 3,12 m<\/li>\n<\/ol>\n\n\n\n

                    Pembahasan:<\/p>\n\n\n\n

                    Diketahui:<\/p>\n\n\n\n

                    m<\/em> = 500 gr = 0,5 kg<\/p>\n\n\n\n

                    F<\/em> = 0,25 N<\/p>\n\n\n\n

                    t<\/em> = 3 s<\/p>\n\n\n\n

                    Ditanya: s<\/em> =\u2026?<\/p>\n\n\n\n

                    Jawab:<\/p>\n\n\n\n

                    Mula-mula, kamu harus mencari percepatan gerak mobil-mobilan setelah didorong adik. Dalam hal ini, gunakan persamaan Hukum Newton 2.<\/p>\n\n\n\n

                    \"\"<\/figure>\n\n\n\n

                    Selanjutnya, gunakan persamaan jarak pada GLBB.<\/p>\n\n\n\n

                    \"\"<\/figure>\n\n\n\n

                    Jadi, jarak yang ditempuh mobil-mobilan setelah adik mendorongnya selama 3 s adalah 2,25 m.<\/p>\n\n\n\n

                    Jawaban: A<\/p>\n\n\n\n

                    Contoh Soal 10<\/strong><\/h2>\n\n\n\n

                    Dua buah balok, yaitu balok P dan balok Q, dihubungkan oleh sistem katrol seperti berikut.<\/p>\n\n\n\n

                    \"\"\/<\/figure>\n\n\n\n

                    Massa balok P dan balok Q berturut-turut adalah 1 kg dan 2 kg. Saat balok Q bergerak ke bawah, balok P bergerak ke kanan. Jika balok P berada di atas meja yang licin dengan massa dan gesekan katrol diabaikan, besar gaya tegangan tali pada balok Q adalah \u2026.<\/p>\n\n\n\n

                      \n
                    1. 9,12 N<\/li>\n\n\n\n
                    2. 8,67 N<\/li>\n\n\n\n
                    3. 12,3 N<\/li>\n\n\n\n
                    4. 3,25 N<\/li>\n\n\n\n
                    5. 6,67 N<\/li>\n<\/ol>\n\n\n\n

                      Pembahasan:<\/p>\n\n\n\n

                      Diketahui:<\/p>\n\n\n\n

                      m<\/em>p<\/sub><\/em> = 1 kg<\/p>\n\n\n\n

                      m<\/em>Q<\/sub><\/em> = 2 kg<\/p>\n\n\n\n

                      Ditanya: T<\/em> =\u2026?<\/p>\n\n\n\n

                      Jawab:<\/p>\n\n\n\n

                      Mula-mula, kamu harus menguraikan komponen gaya yang bekerja pada sistem katrol tersebut.<\/p>\n\n\n\n

                      \"\"\/<\/figure>\n\n\n\n

                      Selanjutnya, kamu bisa meninjau arah gerak baloknya. Untuk balok P, arah geraknya searah dengan sumbu-x. Artinya, tidak ada gerak pada sumbu-y. Dengan demikian:<\/p>\n\n\n\n

                      Balok P<\/strong><\/p>\n\n\n\n

                      \"\"<\/figure>\n\n\n\n

                      Sementara itu, pada balok Q arah geraknya searah sumbu-y. Dengan demikian:<\/p>\n\n\n\n

                      Balok Q<\/strong><\/p>\n\n\n\n

                      \"\"<\/figure>\n\n\n\n

                      Lalu, substitusikan nilai T<\/em> pada persamaan (1) ke persamaan (2), sehingga diperoleh nilai percepatan seperti berikut.<\/p>\n\n\n\n

                      \"\"<\/figure>\n\n\n\n

                      Gaya tegangan tali bisa kamu tentukan dari salah satu balok. Coba kamu ambil balok P.<\/p>\n\n\n\n

                      \"\"<\/figure>\n\n\n\n

                      Ingat, gaya tegangan tali balok Q = gaya tegangan tali balok P.<\/p>\n\n\n\n

                      Jadi, besarnya gaya tegangan tali pada balok Q adalah 6,67 N.<\/p>\n\n\n\n

                      Jawaban: E<\/p>\n\n\n\n

                      Itulah 10 contoh soal Hukum Newton 2 kali ini. Semoga bisa bermanfaat untuk kamu jadikan acuan belajar, ya. Agar belajarmu tambah semangat, yuk buruan gabung Quipper Video. Salam Quipper!<\/p>\n","protected":false},"excerpt":{"rendered":"

                      Hai Quipperian, di artikel sebelumnya Quipper Blog sudah pernah membahas tentang Hukum Newton 2 beserta penerapannya. Apa sih yang kamu ingat dari hukum ini?…<\/p>\n","protected":false},"author":156447303,"featured_media":310694,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[679384877],"tags":[679386726],"ppma_author":[679386823,679386838],"class_list":["post-310690","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fisika","tag-materi-ipa-kelas-8"],"acf":[],"yoast_head":"\nKumpulan Contoh Soal Hukum Newton 2 Lengkap dengan Jawaban dan Penjelasan - 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