Calculating force and work done with 3D vectors

[r.physics]

The question is as follows:

An object of mass m=1 follows the trajectory:

r(t) = 4 ln(t) i + 6t^1/2 j + 2t k

Calculate the force acting on the object and hence find work done between t=1 and t=2.

I know that Force = Mass * Acceleration

Therefore, F(t) = 1 * a(t)

I also know that a(t) = r(t)''

After differentiating I end up with r(t)'' = -4t^-2 i - 3/2 t^-3/2 j

So I've ended up with

F(t) = -4t^-2 i - 3/2 t^-3/2 j

Is that my final answer for the force acting on the object, do I leave it in vector form?

For the work done part, I know that Work Done = Force x Distance

I have force and displacement in vector form but I don't know how to end up with a completely numerical value as I've got a time interval for t, I'm assuming there will be some sort of integration involved.

Can someone please tell me if I have done the first part correctly (calculating the force acting on the object) and if so how can I use this along with the conditions t=1 and t=2 to calculate work done? Thanks!

 

[tommyxu3]

Is that my final answer for the force acting on the object, do I leave it in vector form?

I think so, for the general concept of force contains its direction.
For the remained part, I suggest you try to calculate the tiny work $dW$ with the tiny displacement $d\mathbf{r}$ and then get their sum.

 

[r.physics]

I think so, for the general concept of force contains its direction.
For the remained part, I suggest you try to calculate the tiny work $dW$ with the tiny displacement $d\mathbf{r}$ and then get their sum.

Thankyou for your reply. If I'm understanding what your suggesting correctly, I would now do:

∫dw = ∫ (F ⋅ dr) between the limits of t=2 and t=1

This leaves me with

Δw = [ 4t^-1 i + 3t^-1/2 j + 2 k ]⋅[4 ln(t) i + 6t^1/2 j + 2t k]

With the limits being 2 and 1, I don't think I have done what you meant correctly as when I plug the limits in I end up with a negative answer for work done, any idea where I'm going wrong?

Thanks!