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During an automobile crash test, a 2000kg car is sent toward a cement wall at a speed of 14m/s and is brought to a stop in 0.5s. What was the average force of the car on the wall?

A. 112,000 N
B. 56,000 N
C. 28,000 N
D. 14,000 N

User Ger Groot
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1 Answer

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Final answer:

To calculate the average force during an automobile crash test, the formula F = m \( \Delta v\) / \( \Delta t\) is used, yielding a force of 28,000 N.

Step-by-step explanation:

The subject question inquires about the average force exerted on a wall by a car during an automobile crash test. To calculate this force, we use Newton's second law of motion and the formula for average force, which is F = m \( \Delta v\) / \( \Delta t\), where F is the force, m is the mass of the car, \( \Delta v\) is the change in velocity, and \( \Delta t\) is the time over which the change occurs. Given a 2000kg car going at 14m/s, brought to a stop in 0.5s:

\( F = 2000kg \times (0m/s - 14m/s) / 0.5s \)

\( F = -28,000 kg \cdot m/s^2 \)

The average force of the car on the wall is 28,000 N in the direction opposite to the initial motion of the car, indicated by a negative sign, which is common when indicating opposing forces. So, the correct answer to the question is C. 28,000 N.

User Akilan
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