Sebuah batu 2 kg bergerak pada
kecepatan 6 m/s. Hitung gaya F yang
dapat menghentikan batu itu dalam waktu 7 x 10-4 detik !
A 2 kg brick is moving at a speed
of m/s. How large a force F is needed
to stop the brick in a time of 7 x 10-4 s !
Untuk melihat teori momentum & impuls; hukum
kekekalan momentun; tumbukan, silakan buka masing-masing pembahasannya di Momentum & Impuls; Hukum Kekekalan Momentum ;Tumbukan
To view the theory of momentum and impulse; law of conservation of momentum; collision, please go to the respective discussion in Momentum & Impuls; Hukum Kekekalan Momentum ; Tumbukan
To view the theory of momentum and impulse; law of conservation of momentum; collision, please go to the respective discussion in Momentum & Impuls; Hukum Kekekalan Momentum ; Tumbukan
Penyelesaian:
Pakailah persamaan impuls:
\[\mathrm{ impuls\,yang\,diberikan\,kepada\,batu }=\mathrm{ perubahan\,momentum\,batu}\]
\[F\Delta {t}=mv_{f}-mv_{o}\]
\[F(7x10^{-4})=0-(2\,kg)(6\,m/s)\]
Jadi F = - 1,7 x 104 N. Tanda minus menunjukkan arah gaya
yang menghambat gerak batu tersebut.
Let us solve this by use of the
impulse equation :
\[\mathrm{impulse\,on\,brick}=\mathrm{change\,in\,momentum\,of\,brick}\]
\[F\Delta {t}=mv_{f}-mv_{o}\]
\[F(7x10^{-4})=0-(2\,kg)(6\,m/s)\]
From which F = - 1,7 x 104 N. The minus sign indicates that the
force opposes the motion.