Finding velocity from force and distance

[JKGlover]

Homework Statement

mass=60kg
distance=9.6m

Find velocity

Homework Equations

force=mass*acceleration

The Attempt at a Solution

I multiplied mass by accelleration(9.8) and got 588. Thats the force.

What do I do with force and distance to get velocity?

 

[Angry Citizen]

Could you be more specific? Are you dropping a mass from a given height and trying to find the velocity upon impact?

The most important part of physics is learning how to state a problem. This may be your issue.

 

[JKGlover]

Yes. I'm trying to find the velocity on impact.

Sorry for the inconvience

 

[Angry Citizen]

Do you have to use force? I'd use the principle of conservation of energy for this problem. It practically screams for it.

 

[JKGlover]

What is that?

I haven't run into that yet.

 

[Angry Citizen]

Oh.

Hmm, well, this is the advanced physics section, so I'd assumed you had.

Are you in calc-based physics? If so, you might use f=ma and use the fact that a=dv/dt=(dv/ds)(ds/dt), and recognize that ds/dt is v. This should yield an expression for v using only distance as your independent parameter.

 

[JKGlover]

Yes I'm in calc-based physics.

Because it's being dropped from zero velocity wouldn't the acceleration be gravity(9.8)?

If so, are you saying that I should solve for ds/dt?

 

[Angry Citizen]

No.

If a=9.8, then you need to integrate the expression dv*v=a*ds to find an expression for v, where ds is a differential arc length (luckily in your case you can assume that the arc has no curvature, and so your line integral breaks down into just an ordinary integral as you're familiar with - in other words, s is just the height). You are just using the chain rule as described in post #6 to create an expression that uses the variables you know, i.e. distance and acceleration, to find one you don't, i.e. velocity.

The resulting equation should be in your book, but now you know how it's derived.

 

[JKGlover]

Thank you so much I understand now. But one remaining question.

in"dv", does that mean distance*velocity or the velocity traveled in the distance?

 

[Angry Citizen]

That's actually a differential unit of velocity. Like in calculus the expression "dx" is a differential length in the x direction. Same thing here. Since velocity is always changing, you want to isolate an infinitesimally small unit of velocity such that the velocity is "constant" over that infinitesimally small period of time (or in this case, distance, since the differential is dv/ds). This allows you to sum up all the infinitesimally small "chunks" of velocity - which is what an integral does.

The extra v term is used just like you would if you were integrating x*dx.

Sorry if this is confusing in any way. I realize now that maybe this derivation may be a little much. But it's EXTREMELY HANDY to know how these equations of motion are derived. I didn't really learn kinematics until I knew the derivations.

 

[JKGlover]

ok thanks