What average force is it brought to rest?

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A 1000 kg car traveling at 14 m/s comes to a stop upon hitting a cement wall in 0.08 seconds. The average force required to bring the car to rest is calculated using the formula F = mv/t, resulting in 175,000 N. This force corresponds to an average acceleration of 175 m/s², which is nearly 18 times the acceleration due to gravity. The calculations confirm the significant impact force experienced during the crash test. Understanding these forces is crucial for assessing vehicle safety in crash scenarios.
missashley
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Homework Statement



Auto companies frequently test the safety of automobiles by putting them through crash tests to observe the integrity of the passenger compartment. If a 1000. kg car is sent toward a cement wall with a speed of 14 m/s and the impact brings it to a stop in 8.00 * 10^-2 seconds, with what average force is it brought to rest?

Homework Equations



F = mv/t

The Attempt at a Solution



F = mv/t = (1000*14)/(8e-2) = 175000 N
 
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missashley said:
F = mv/t = (1000*14)/(8e-2) = 175000 N

There you have it! The average acceleration during the simulated crash is the change in velocity (14 - 0 m/sec) divided by the time interval over which it occurred (0.08 sec) or 175 m/(sec^2) , which is almost 18 times the acceleration of gravity. The average force on the car is then its mass times that average acceleration or

175 m/(sec^2) · 1000 kg = 175,000 N (or almost 18 times the car's weight).
 
Thats right.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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