<span>Step 1 -- determine the acceleration of the 200-g block after bullet hits it
a = (coeff of friction) * g
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
a = 0.400*9.8
a = 3.92 m/sec^2
Step 2 -- determine the speed of the block after the bullet hits it
Vf^2 - Vb^2 = 2(a)(s)
where
Vf = final velocity = 0 (since it will stop)
Vb = velocity of block after bullet hits it
a = -3.92 m/sec^2
s = stopping distance = 8 m (given)
Substituting values,
0 - Vb^2 = 2(-3.92)(8)
Vb^2 = 62.72
Vb = 7.92 m/sec.
M1V1 + M2V2 = (M1 + M2)Vb
where
M1 = mass of the bullet = 10 g (given) = 0.010 kg.
V1 = velocity of bullet before impact
M2 = mass of block = 200 g (given) = 0.2 kg.
V2 = initial velocity of block = 0
Vb = 7.92 m/sec
Substituting values,
0.010(V1) + 0.2(0) = (0.010 + 0.2)(7.92)
Solving for V1,
V1 = 166.32 m/sec.
Therefore the answer is (B) 166 m/s!</span>
Answer:
I agree with Katelyn's statement
Explanation:
Netwon's first law of motion states that an object will remain at rest or in uniform motion except acted upon by an external (unbalanced) force.
When balanced forces are acting on an object in motion, the object maintains its speed and direction.
This means that balanced forces do not change the speed and direction of an object in motion. We can say that the object is in equilibrium. It either remains at rest or in uniform motion.
Therefore, I agree with Katelyn's statement