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ssm mmh A 1580-kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 50.0 m?

3560 $\mathrm{N}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

University of Michigan - Ann Arbor

University of Sheffield

McMaster University

Lectures

04:16

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Newton's second law tells us that the net force it's not, is equal to the match off the car. Times declaration. We know the math of the car is equals. True, one, 580 kilo grams. It isn't with a play about acceleration. Now we have to calculate what is the relation of this car? How can we do that? Well, we have the variation in the velocity and we have the displacement that happen why August velocity was creating. But we don't know what is the time interval in between these three events. So when you don't have time, we used to return his equation, which in the following the final Velocity Square is equal to the initial velocity squared quest to times declaration, times displacement. So in this question, the final velocity is close to zero. The initial velocity vehicles to 15. Um, we have two times declaration that we want to calculate times the displacement off 50 meters. Then we go minus two times declaration times 50 whose egos to 15 square miners. Because we've sent this turn to the other side. Then we get it that the acceleration is because two minors it's been squared divided by two times 50. We just sent both the truth and the 50 to another side. The hiding. Then acceleration is equal to minus if squared, divided by 100. And this is equally true. Minus 225 divided by 100 which is minors to going 25 meters per second. Squared the observations negative because the velocity is being reduced. Then we can plug in the velocity that we have just calculated. Even you turn second law equation to get the following that horse that's next is he goes to 150 times minus two point point five. This gives us a net force approximate minus 3560 new toes.

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