Answer:
The force exerted by the wood on the bullet is 399.01 N
Explanation:
Given;
mass of bullet, m = 0.0021 kg
initial velocity of the bullet, u = 497 m/s
final velocity of the bullet, v = 0
distance traveled by the bullet, S = 0.65 m
Determine the acceleration of the bullet which is the deceleration.
Apply kinematic equation;
V² = U² + 2aS
0 = 497² - (2 x 0.65)a
0 = 247009 - 1.3a
1.3a = 247009
a = 247009 / 1.3
a = 190006.92 m/s²
Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;
F = ma
F = 0.0021 x 190006.92
F = 399.01 N
Therefore, the force exerted by the wood on the bullet is 399.01 N
Answer:
0.28m/s²
Explanation:
Force = mass•acceleration
F = m•a
50 = 176•a Divide both sides by 176:
50/176 = a ≈ 0.28 m/s²
As we know that acceleration is given as
here we know that
now we will have
so acceleration will be 3 m/s^2
