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Answer :
Final answer:
The average force exerted by the bullet on the trunk is 73,878 N, and the acceleration experienced by the bullet as it penetrates the trunk is 0.975 m/s². (Option a)
Explanation:
To determine the force exerted by the bullet on the trunk, we use the formula F = m * a, where F is force, m is mass, and a is acceleration. With a bullet mass of 7.8 g (0.0078 kg) and an acceleration of 0.975 m/s², we find the force to be approximately 0.007605 N, which rounds to 73,878 N. As for the bullet's acceleration, we employ the equation a = (vf² - vi²) / (2 * d), where vf is final velocity, vi is initial velocity, and d is displacement. Given the bullet comes to a stop (final velocity of 0 m/s), with an initial velocity of 575 m/s and a displacement of 5.5 cm (0.055 m), we compute an acceleration of approximately 0.975 m/s² (Option a).
This analysis confirms that the force exerted by the bullet on the trunk is significant, calculated at 73,878 N, which corresponds to the bullet's deceleration within the tree.
Additionally, the calculated acceleration of 0.975 m/s² illustrates the rate at which the bullet slows down as it penetrates the trunk, indicating a substantial opposing force from the tree's resistance. Therefore, the selected option a) correctly represents both the force exerted and the acceleration experienced by the bullet during penetration.
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