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Example 2.1. on pg 30
Example 2.1. on pg 30

... A1500kg car moving at 15m/s collides with a utility pole and is brought to a rest in .3sec. Find the average force exerted on the car during the collision. F = (1500kg)(0m/s-15m/s)/.3sec = -75000N The negative sign means the force is in the opposite direction of the car’s velocity. Go over example 6 ...
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Exam 2 solutions - BYU Physics and Astronomy

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During a relay race, runner A runs a certain distance due north and

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6.8 Mb - Todd Satogata

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Newton`s Laws

... Each force is the same size. For every action, there is an equal ... (equal!). The fact that the firefly splatters only means that with its smaller mass, it is less able to withstand the larger acceleration resulting from the interaction. ...
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Student Exam Review

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Chapter 4 Forces and Newton’s Laws of Motion Conclusion

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Experiments Involving Static Equilibrium

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Note that in the following three figures, which show

... We learned earlier that if an object is at rest, then the total of all the external forces acting on it must be zero. If we are also considering the possibility of rotational motion, we must add a second condition for a body at rest: The net torque of all the external forces acting on the body, wit ...
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