Force Applied to an Object with friction

1. Feb 20, 2013

EnDev

1. The problem statement, all variables and given/known data

2kg block moving across a flat surface with a friciton coefficient of .2

a. If 10N force applied what is acceleration

b How fast will the block move if the force is applied for 5seonds

c. how far will the block have traveled after 5 seconds

2. Relevant equations

Not sure...

3. The attempt at a solution

So after applying friction i get a net force of 6.08 Newtons. plug that back into F = M*A
giving an accelration of 3.04 m/s2 , Not sure if i Square root the 3.04 to get 1.74 m/s so that i get the acceleration. then if the force is applied for 5 seconds then 1.74 * 5 to get 8.7 m/s

and after 5 seconds the block would travel 40.35 meters??

2. Feb 20, 2013

tms

Why would you take the square root?

You don't say what the initial velocity is, which is needed to solve the second and third parts of the problem.

3. Feb 20, 2013

EnDev

Ok, scratch that then, but there would be no velocity as the object is at rest until the force is applied.

4. Feb 20, 2013

haruspex

This is five seconds after applying the force, so it has been accelerating at 3.04 m/s2 for 5 seconds. An acceleration of 3.04 m/s2 means that each second it travels... how much faster?

5. Feb 20, 2013

EnDev

so 3.04 * 5 = 15.2 m/s the object is traveling after 5 seconds?

or would 15.2 be the distance it traveeled after 5 seconds?

6. Feb 20, 2013

EnDev

Wait, it would still travel at 3.04 m/s right? no matter the distance?

and after 5 secs it would travel 15.2 meters

7. Feb 20, 2013

haruspex

Yes.
You seem quite confused about distance versus speed versus acceleration.
If it accelerates at 3.04 m/s2 then each second it gets 3.04m/s faster. "3.04 metres per second, per second." After 1 second it's speed is 3.04 m/s. After 2 seconds it's 6.08 m/s... You might keep track of it better if you include the units as though they're algebraic variables: 3.04 m/s2 * 5 s = 15.4 m/s; m/s2 * s = m/s.
What kinematic equation do you know relating distance to time and to constant acceleration?

8. Feb 20, 2013

EnDev

would it be d = v^i*t + 1/2*a*t^2 ???

9. Feb 20, 2013

EnDev

so the block would have traveled 35.60 meters?

10. Feb 20, 2013

tms

Not quite. You have
$$x = \frac{1}{2} at^2 = \frac{1}{2} (3.04)(5)^2 = \:?$$
assuming the initial position and velocity were zero.

11. Feb 20, 2013

tms

You need to put things like that in the statement of the problem. It is best to quote the problem exactly, rather than try to paraphrase it, because it is possible to misinterpret a problem (you didn't here, but it can happen). If you quote exactly, others might catch the misinterpretation.

12. Feb 20, 2013

EnDev

I apologize for that.

but as for the equation i get 38meters.

Im gonna work on the next one to be sure that ive got this down.

thanks for the help!

13. Feb 20, 2013

tms

No need to apologize.
That's what I got.

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