Solving problems involving forces

[afifbaha]

Homework Statement

A loaded barge has a mass of 1.5exponent6 kg and its traveling at 3 metres per second. If a tugboat applies an opposing force of 1.2exponent4 N for 10s

What is it final velocity?
How long does it take to stop

Homework Equations

F = ma
F = m(v-u)/t

The Attempt at a Solution

 

[CWatters]

Please post your attempt at a solution.

 

[afifbaha]

Please post your attempt at a solution.

Ive subtituted all the values given in the formula. From then, i couldn't figure out the next step

 

[CWatters]

It's best NOT to substitute values at the outset. The problem asks for the final velocity. Try rearranging the second equation to give an equation for the final velocity. Show your working.

 

[afifbaha]

It's best NOT to substitute values at the outset. The problem asks for the final velocity. Try rearranging the second equation to give an equation for the final velocity. Show your working.

So can i substitute F on the second equation with the value of N given in the question. It says that it is an opposing force. So i have to do F-N = m(v-u)/t. Right? Then i will have 2 unknowns F and V. What now?

 

[CWatters]

No it's easier than that. The "N" in the question just stands for Newtons the unit of force not another variable.

|F| = 1.2 * 10⁴ Newtons

Actually you need to think about the sign.

 

[haruspex]

The second question is a bit unclear. I think it means, how long will it take to stop if the same force continues?

 

[TomHart]

I think haruspex is right that the question is asking how long it will take to stop if the same opposing force from the tugboat continues. The reason is because I think this problem is intended to be equivalent to an object sliding on a frictionless surface. In reality, a barge and a tugboat is not that situation because of the varying resistance of the water. So I think a barge and tugboat is really not a good choice for a problem of this nature.

I also had to look up the equation F = m(v - u)/t because I didn't know what it meant. Maybe I am revealing my age, but I remember V = Vo + at, which would rearrange to F = m(V-Vo)/t. But afifbaha, please use the equation that you listed and are familiar with: F = m(v-u)/t

So as CWatters said, try rearranging the equation to solve for the unknown.
In the first part of the problem, the final velocity is the unknown. In the second part of the problem, the final velocity is given and time is the unknown.

And like CWatters pointed out, 'N' is not a variable; it is a symbol for Newtons - a unit of force, like 'lb' for pounds.

Sorry to be so wordy, but this is my very first post after my introduction.
~ inept new guy

 

[Said Ahmad]

I suggest to use momentum & impuls equation

 

[Said Ahmad]

Actually your second equation is the simple impuls equation,
You can solve the first question with your second equation, just subtitute and solve...

 

[CWatters]

I think we should let the OP attempt the problem again offering more help.

 

[afifbaha]

I knew that N is Newtons. What i was saying was for F in F=ma, the F is net force which means it has to be subtracted with any other forces. In this case it has to be subtracted with the opposing force right? That is what i meant.

 

[CWatters]

Ok sorry about that.

As far as I can see the tug is the only force acting on the barge - at least no other forces are stated in the problem.

 

[afifbaha]

Ya that's what i was figuring out. Anybody else could help?

 

[haruspex]

Ya that's what i was figuring out. Anybody else could help?

CWatters is saying (and I agree) that you should consider the tug's arresting pull as the only horizontal force on the barge. Of course, in reality, there would be drag from the water, but that would be very small in comparison.