A 6 g bullet leaves the muzzle of a rifle with a speed of 333.3 m/s.
What constant force is exerted on the bullet while it is traveling down the 0.9 m length of the barrel of the rifle?
Answer in units of N
To find the constant force exerted on the bullet while it is traveling down the length of the barrel, we can use the principle of conservation of energy.
The initial kinetic energy of the bullet can be calculated using the formula: KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass of the bullet, and v is the velocity of the bullet.
Given:
m = 6 g = 0.006 kg (since 1 g = 0.001 kg)
v = 333.3 m/s
Plugging in these values, we can calculate the initial kinetic energy:
KE = (1/2)(0.006 kg)(333.3 m/s)^2
= (1/2)(0.006 kg)(111108.89 m^2/s^2)
= (0.003 kg)(111108.89 m^2/s^2)
= 333.3267 J (joules)
Since there is no change in height or speed, we can assume that all the initial kinetic energy is converted to work done by a constant force over the distance traveled (0.9 m).
Work done is given by the formula: W = Fd, where W is the work done, F is the force applied, and d is the distance traveled.
Therefore, we can set up the equation:
333.3267 J = F * 0.9 m
To find F, we can rearrange the equation:
F = 333.3267 J / 0.9 m
= 370.363 N (Newtons)
Therefore, the constant force exerted on the bullet while it is traveling down the 0.9 m length of the barrel is approximately 370.363 N.
Vf^2=Vi^2+2ad where a=force/mass
solve for force