2. A rifle bullet of mass 7.86 grams starts from rest and exits from the 2.188feet barrel at 4320km/hr. Find the force exerted on the bullet, assuming it to be constant, while the bullet is in the barrel.
Answers
Explanation:
2. A rifle bullet of mass 7.86 grams starts from rest and exits from the 2.188feet barrel at 4320km/hr. Find the force exerted on the bullet, assuming it to be constant, while the bullet is in the barrel.
Given info : A rifle bullet of mass 7.86 grams starts from rest and exits from the 2.188 feet barrel at 4320km/hr.
To find : The force exerted on the bullet, assuming to be constant, while the bullet is in barrel.
solution : distance covered by bullet in barrel, s = 2.188 ft = 0.67 m
initial velocity of bullet , u = 4320 km/h = 4320 × 5/18 = 1200 m/s
final velocity of bullet, v = 0.
using formula, v² = u² + 2as
⇒0 = 1200² + 2a(0.67)
⇒a = 1200²/1.34 = 1,074,626.87 m/s²
Now the force exerted on the bullet , F = ma
= 7.86 g × 1,074,626.87 m/s²
= 7.86 × 10¯³ kg × 1,074,626.87 m/s²
= 8446.5672 N
Therefore the force exerted on the bullet is 8446.5672 N
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