# Newton's first and secon law

1. Mar 4, 2004

### Beretta

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 dont know how to get the accelration since I dont have the time t, so I can calculate the mass.

2. Mar 4, 2004

### Tom McCurdy

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.

3. Mar 5, 2004

### HallsofIvy

Staff Emeritus
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.

4. Mar 5, 2004

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