Finding displacement using net work?

[ooohffff]

Homework Statement

The question is: A 2 kg particle is moving in the positive x direction with the speed of 5 m/s. As it passes the origin, a force F = (30 N - 2N/s*t)i is applied to it. Where does the particle come to a stop?

Homework Equations

F=ma
W = integral F(x)dx
K=.5mv^2

The Attempt at a Solution

K1 = .5(2)(5)^2 = 25 J
K2 = .5(2)(0)^2 = 0 J
W = K2-K1= -25 J

My main issue is I'm not sure how to rewrite the force equation in terms of x, since I can use it in the formula Work = integral F(x)dx from 0 to x to solve for x.

Thanks!

 

[cnh1995]

I deleted my previous post. I think the suggestion was not correct. You need to first write v as a function of t and find the value of t for which v=0. Then you should write displacement x as a function of time and put the value of t to get the displacement. You'll need differential equations for this. What is the relation between v, F and t?

 

[ooohffff]

F=ma=m(v/t)

I deleted my previous post. I think the suggestion was not correct. You need to first write v as a function of t and find the value of t for which v=0. Then you should write displacement x as a function of time and put the value of t to get the displacement. You'll need differential equations for this.

So since F=ma, I divided the force vector by 2kg, giving me an acceleration vector of a=(15-t)i

Taking the integral of that, I would get a velocity vector of v = (15t - .5t^2 + 5)i. Set v=0, then I get t=30.33 s.

Does that look right so far?

 

[cnh1995]

F=ma=m(v/t)So since F=ma, I divided the force vector by 2kg, giving me an acceleration vector of a=(15-t)i

Taking the integral of that, I would get a velocity vector of v = (15t - .5t^2 + 5)i. Set v=0, then I get t=30.33 s.

Does that look right so far?

Yes.

 

[ooohffff]

Yes.

Ok, so the final result doesn't look right...So taking the integral of v, I get x=[-1.5t^3 + 7.5t^2 + 5t]i . Plugging in 30.33 s, I get x = -384803.29

 

[cnh1995]

Ok, so the final result doesn't look right...So taking the integral of v, I get x=[-1.5t^3 + 7.5t^2 + 5t]i . Plugging in 30.33 s, I get x = -384803.29

Check the integration again. The answer comes out to be positive.

 

[ooohffff]

Check the integration again. The answer comes out to be positive.

Oops, you're right, it should be (-1/6)t^3 not (-3/2)t^3. And the answer comes out to 2400 m

 

[cnh1995]

Oops, you're right, it should be (-1/6)t^3 not (-3/2)t^3. And the answer comes out to 2400 m

Right. 2400.82...

 

[ooohffff]

Right. 2400.82...

Yup! I only need 2 sig figs, though. Thank you so much for your help!

 

[cnh1995]

Thank you so much for your help!

No problem! And welcome to Physics Forums!