The force exerted on a bullet of mass 1.0 g by the block is -970 N
Explanation:
The motion of the bullet is a uniformly accelerated motion, so we can calculate its acceleration by using the suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
Here we have:
u = 550 m/s
v = 420 m/s
s = 6.5 cm = 0.065 m
Solving for a,
[tex]a=\frac{v^2-u^2}{2s}=\frac{420^2-550^2}{2(0.065)}=-9.7\cdot 10^5 m/s^2[/tex]
Now we can find the force exerted on the bullet, which is given by
[tex]F=ma[/tex]
where
m is the mass of the bullet. Here it is not given, so I'll assume that its value is
m = 1.0 g = 0.001 kg
a is the acceleration
And substituting, we find
[tex]F=(0.001)(-9.7\cdot 10^5)=-970 N[/tex]
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