Newtons Laws of Motion Question

AI Thread Summary
The discussion revolves around a physics problem involving the acceleration and time of a Tyrannosaurus rex's torso upon impact with the ground. Initially, the user incorrectly calculated the torso's upward acceleration by not accounting for the weight of the T-Rex, leading to an erroneous result. After correcting the calculation by subtracting the weight from the net force, the correct upward acceleration was found to be 59.39 m/s². The user then faced difficulties with part b of the problem, specifically in calculating the time taken to come to rest after impact, indicating confusion over the variables and the equations needed for the solution. Clarifications were provided regarding the need for separate equations to determine the velocity upon impact and the time to decelerate.
laurenflakes
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Homework Statement


Paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced a net force of approximately 256,700 N acting on its torso when it hit the ground. Assume the torso has a mass of 3710 kg.

(a) Find the magnitude of the torso's upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about 7g.)

(b) Assuming the torso is in free fall for a distance 1.39 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?


Homework Equations



F = (m)(a)

The Attempt at a Solution



Immediately it appeared to me that by using the equation that states that the sum of the net forces is equal to the mass of the object in kilograms multiplied by the acceleration of the object. I rearranged the equation to solve for acceleration as such:

256700 N / 3710 kg = 69.19 m/s^2

However, this answer is incorrect.. I don't see what I am doing wrong? Help??!? :(
 
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Nevermind! I got the answer :)

What I was doing wrong was not taking into account the affect of the weight on the Net force.

By determining the weight of the T-Rex in Newtons (3710)(9.8) = 36358 and subtracting this number from the original net force 256700 - 36358 = 220342 N then taking that number and dividing by the mass 220342 N/3710 kg I was able to get the correct acceleration which was 59.39 m/s^2 :)

Still stumped on part b though. Now that I know the acceleration I tried using the equation v(final) ^2 = v(initial)^2 + 2(a)(d). With my final velocity being 0 (the trex coming to rest) and using the acceleration calculated in part a and the distance given to calculate v-final... however the answer that I get produces the wrong answer for the time... any suggestions?
 
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Could you show your working? I'm a bit suspicious of the variables that you are using - they seem to be mixed up, especially d, which according to the question is the distance through which it freely falls before it gets decelerated, and not the distance traveled during the deceleration.
You should have two sets of equations, one to determine the velocity of the torso upon impacting the ground, and the second to determine the time taken to bring it to rest.
 
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