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The answer is 2.93m (to 2 dp)
Answer:
a) 4 289.8 J
b) 4 289.8 J
c) 6 620.1 N
d) 411 186.3 m/s^2
e) 6 620.1 N
Explanation:
Hi:
a)
The kinetic energy of the bullet is given by the following formula:
K = (1/2) m * v^2
With
m = 16.1 g = 1.61 x 10^-2 kg
v = 730 m/s
K = 4 289.8 J
b)
the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest) Ki = 0, therefore:
W = ΔK = Kf - Ki = 4 289.8 J
c)
The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).
If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:
W = F * L
Where F is the net force and L is the length of the barrel, that is:
F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N
d)
The acceleration can be found dividing the force by the mass:
a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2
e)
The force will have a magnitude equal to c) and direction along the barrel towards the exit
