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A 950 -kg car strikes a huge spring at a speed of 25 $\mathrm{m} / \mathrm{s}$ (Fig. 41$)$ , compressing the spring 5.0 $\mathrm{m}$ . (a) What is the spring stiffness constant of the spring? (b) How long is the car in contact with the spring before it bounces off in the opposite direction?

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a. $$2.4 \times 10^{4} N / m$$b. $$0.62$$

Physics 101 Mechanics

Chapter 14

Oscillators

Motion Along a Straight Line

Motion in 2d or 3d

Periodic Motion

University of Michigan - Ann Arbor

University of Washington

Hope College

McMaster University

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

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In physics, an oscillation is the repetitive variation, typically in time, of some measure about a central value or between two or more different states. The oscillation may be periodic or aperiodic.

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A 950-kg car strikes a hug…

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so for party. We're trying to find the spring constant so we can use the law of conservation of energy, he said. One equals east of two or essentially, the total amount of energy does not change and we can say 1/2 and the sub one squared, Um simply equals 1/2 K except two squared. Given that there's no, uh, there's no potential energy in the beginning and there is no kinetic energy after the spring is fully stretched And so we can say that this is gonna be and V one squared equals K X, some two squared. And at this point, weekend US calculate case Okay, with the equal to M v sub one squared, divided by except two squared and this is gonna equal the mass of the car. So 950 kilograms times 25 meters per second squared 25 meters per second, quantity squared, divided by 5.0 meters. Quantity squared, and we find that the spring constant is going to be 2.375 times 10 to the fourth, a Newtons per meter and so this will be our final answer for a part for part B. However, we need to find the this here. They're saying that the car, how long will the car be in contact with spring. So from the time it takes to reach them, not from the maximum displacement to the equilibrium position is always equal to half the period. So in this case, the car will be in contact with the spring for half of the period. Uh, we can say that, um, Max, displacement to equilibrium position, uh, half of the period. So we can say that if we wanted to solve half of tea would be equal to 1/2 of to pi times the square root of the mass divided by the spring constant. And this would be equal to pi times the square root of 950 kilograms. And then this will be divided by the spring constant of 2.3775 times 10 to the fourth Newtons per meter. And we find that the the the time taken with the 0.63 seconds. So that's how long the car will be in contact with the spring. This would be a final answer for part B. That is the end of the solution. Thank you for watching

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