A 12 gram bullet is accelerated from rest to a speed of 700 m/s as it travels 20 cm in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force?
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Answered by
12
solution
from 3rd equation of motion
2as=v^2 - u^2
putting value
2a (0.20)=700^2 - 0^2
0.4a= 490000
Divide both side by 0.4
you will get
a=1225000m/s^2
Now we will find force from newton 2nd law of motion .
F=ma
putting the value in equation
F=0.012kg * 1225000m/s^2
F=14700N
from 3rd equation of motion
2as=v^2 - u^2
putting value
2a (0.20)=700^2 - 0^2
0.4a= 490000
Divide both side by 0.4
you will get
a=1225000m/s^2
Now we will find force from newton 2nd law of motion .
F=ma
putting the value in equation
F=0.012kg * 1225000m/s^2
F=14700N
Answered by
4
initial velocity of bullet = 700 m/s
finally when bullet comes to rest , velocity = 0 m/s
distance covers during this = 20 cm
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use kinematics equation ,
V² = u² + 2aS
here,
u = 700 m/s
V = 0 m/s
S = 20cm = 20/100 m = 1/5 m
0 = (700)² + 2 × 1/5 × a
- 490000= 2/5 × a
a = - 490000 × 5/2 m/s²
a = - 1225000 m/s²
F = mass of bullet × acceleration
= 12 × 10-³ × (-1225000)
= 14700000 × 10-³
= 14700N
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