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In outer space, a constant net force of magnitude 140 $\mathrm{N}$ is exerted on a 32.5 $\mathrm{kg}$ probe initially at rest. (a) What acceleration does this force produce? (b) How far does the probe travel in 10.0 $\mathrm{s} ?$

see solution.

Physics 101 Mechanics

Chapter 4

Newton's Laws of Motion

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Cornell University

Rutgers, The State University of New Jersey

Hope College

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all right. And this problem, You're looking at a probe in outer space and so are pro outer space has a mass of 32.5 killer cramps. Here's my probe. Mass is equal to 32.5 kilograms, and there is a constant net force of magnitude 140. Newtons applied onto this pro. I'm just choosing to make it go to the right for the sake of choosing of direction. It's not giving in our problem, and in this problem, our probe is initially at rest. And so part eh of this problem is asking us for the acceleration that the force produces and to solve for acceleration. We're going to use Newton's second law, which says that the sum of the forces is equal to mass times acceleration. And so, since we're given Mass and were given force, I'm going to plug those things into this equation. Our net total force on this is 140 Newtons, and I know this is my nut force because there are no other forces that would apply on the probe, especially since this is an outer space. There's no friction. There's no gravity that we have to worry about in this problem. Just the one force our mass is 32.5 and our acceleration is what we're solving for. Divide both sides by 32.5, which gives us our acceleration on one side and 140. Divided by 32.5 is equal to 4.31 and that's meters per second squared. Um, in the direction of that acceleration since acceleration is a vector is whatever direction the force was applied in. And so, in my case to the right for Part B, we want to figure out how far the pro is going to travel in 10 seconds. So let's move our acceleration answer over here and in part B. You want to know how far? And so for this problem, we're going to use a distance equation that you've probably learned in Cana Matics. A cz d equals the initial velocity of an object times the time plus 1/2 times the acceleration times T squared, which is the time squared. And for this problem we're solving for D RV, I our initial velocity is given a zero because it says this is at rest. Initially, Her time is 10 seconds. I mean plus 1/2 a is 4.31 We found that in part a anti again is 10 squared so zero times 10 just cancels out with zero and 1/2 times 4.31 times 10 squared is equal to 215 meters.

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