# Force problem with velocity

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1. May 21, 2015

### Q7heng

1. The problem statement, all variables and given/known data
Hi, on my recent test there was a question stating 2 objects, one with a mass of m, and another with a mass of 4m, are pushed with equal forces on a frictionless track. The question asked which object will reach the end first and I said they will reach at the same time because there is no friction. But the right answer is the smaller mass will reach there first. I don't know how this works into everything in the equations of physics and how force, mass and velocity (not acceleration) is related. Help please!

2. Relevant equations
F=ma

Thanks a lot

2. May 21, 2015

### haruspex

You know how force relates to mass and acceleration. Which mass will have the greater acceleration?
How does velocity depend on acceleration?

3. May 21, 2015

### Q7heng

Shouldn't acceleration be 0 in this case because there is no friction on track?

4. May 21, 2015

### haruspex

I have no idea why you would think that.
Acceleration results from a nonzero net force. If there is an applied force and nothing to oppose it then there will be acceleration.
Indeed, there is less likely to be acceleration when there is friction.

5. May 22, 2015

### CWatters

That would be the case if they were coasting down the track. However the question as written in post #1 states they are being pushed down the track by a force.

6. May 22, 2015

### Jeremymu1195

∑Fx=ma (Newton's Second Law)

F(Push)=ma The only force acting in the direction of motion is the push.

We can also write Newton's second law for the y-direction, but it doesn't tell us anything. ∑Fy=N-mg=0 →N=mg.

Back to the x-direction, we found that a=F(Push)/m by solving Newton's second law in the x-direction for a.

Now we can use kinematics. Since distance is implicitly given, I would use the equation,

v2=2a(xf-xi)=2aΔx

So,

v=√2aΔx or v=√2(F/m)Δx

since m is on the bottom, for a given Δx (i.e. the length of the track) v gets larger as m gets smaller