Work Kinetic energy using variable force

[kjamha]

Homework Statement

The position of an object of mass 2 kg is changing as a function of time by the formula x(t) = 4t^3-6t. (a) find the work done by the forces acting on the particle between t=1 and t=3s. (b) What is the object's velocity when the force on it is zero?

Homework Equations

is my work for part (a) correct?
I do not know how to approach b.

The Attempt at a Solution

I took the derivative of the equation
12(t)^2 - 6 at t=1s and t=3s
to find vi and vf (I have vi = 6 m/s and vf = 102 m/s).
Then I used W = change in KE
1/2 x 2(102^2) - 1/2 x 2 (6^2) = 10,368 J.
Is that correct?
I'm not sure how to start part b

 

[PeroK]

And the time derivative of velocity is?

 

[kjamha]

Are you asking for the acceleration?

 

[PeroK]

Are you asking for the acceleration?

Yes. Acceleration. Does that help for part b?

Your answer to a) is correct.

 

[gneill]

Homework Equations

is my work for part (a) correct?
I do not know how to approach b.

kjamha, your Relevant equations aren't equations. You should place equations that you think are relevant to the type of problem being solved in that portion of the template.

 

[kjamha]

taking the derivative, I would get acceleration = 24t
if F = 0, then 0 = 24t and t would have to be 0
find the velocity at time 0
12(t)^2 - 6 = -6
is the answer -6 m/s?

 

[kjamha]

kjamha, your Relevant equations aren't equations. You should place equations that you think are relevant to the type of problem being solved in that portion of the template.

ok - I read it quickly and thought it said relevant questions.

 

[PeroK]

taking the derivative, I would get acceleration = 24t
if F = 0, then 0 = 24t and t would have to be 0
find the velocity at time 0
12(t)^2 - 6 = -6
is the answer -6 m/s?

Yes, that's correct.

 

[kjamha]

Yes, that's correct.

Thank you!