Calculating Mass Using Newton's Laws

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Homework Help Overview

The problem involves calculating the mass of a particle using Newton's laws, given its initial speed, the net force acting on it, and the distance over which it comes to a stop.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to calculate acceleration without knowing the time. Some participants provide different methods to find acceleration and mass, using kinematic equations and force relationships.

Discussion Status

Multiple approaches to the problem have been presented, with some participants offering calculations based on kinematic equations while others suggest alternative methods. There is no explicit consensus on the best approach, but various lines of reasoning are being explored.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available or the methods they can use. Some assumptions about the direction of forces and the nature of motion are also being discussed.

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A particle of mass m is traveling at an initial speed v0=25.0m/s.
When a net force of 15.0N acts on it, it comes to a stop in a distance of 62.5m. What is m?

I don't know how to get the accelration since I don't have the time t, so I can calculate the mass.

May you please help me out?
 
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Answer... Hopefully

Alright let me see if I can help you out at all

well since V^2=Vo^2+2ax
V^2=0
Vo=25
x=62.5


Therefore 0=25^2+2a62.5
Therefore -625=2a62.5
Therefore a=-5m/s^2

Then I think you can do

F=MA

15=5*M
m=3 kg then I belive.
 
Here's another way to do it, admittedly harder because it doesn't assume that formula, V^2=Vo^2+2ax, Tom McCurdy gave you.

The force is -15N so, with mass m, the acceleration is a= -15/m.
(I'm taking the force to be negative since it stops the motion.)

The initial speed is 25 so, at any time t, the speed is
25- (15/m)t.

The distance moved in time t is 25t- (15/2m)t2.

When the object stops, the speed is 25- (15/m)t= 0 and the distance is 25t- (15/2m)t2= 62.5.

You now have two equations to solve for m and t. Since you were specifically asked for m, it might be simplest to rewrite
25-(15/m)t= 0 as (15/m)t= 25 so t= (25/15)m= (5/3)m. Now put that into the second equation: 25(5/3)m- (15/2m)(25/9)m2= 62.5 which is the same as (125/3)m- (125/6)m= (125/6)m= 62.5 kg.
 
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