1. Feb 18, 2008

jj8890

1. The problem statement, all variables and given/known data
A force F = axÎ+byj , where a = 1.5 N/m and b = 1.7 N/m, acts on an object as the object moves in the Î direction along the x- axis from the origin to c = 4.1 m. Find the work done on the object by the force. Answer in units of J.

Then j is supposed to have a hat on it too and the Force (F) is a vector with the arrow over it.

2. Relevant equations
3. The attempt at a solution

I am pretty sure that I am supposed to take the dot product of the force and the displacement but how do you take the dot product of the force and the displacement when the displacement is only in the i direction? There is no j portion of it.

I would appreciate a little help getting started with this. Thank you

2. Feb 18, 2008

belliott4488

Sure there is! How large is the j component of a vector that points in the i direction? To put it another way: you know your displacement vector is D = 4.1i, right? So if you set D = 4.1i = xi + yj ... what must be the values of x and y?

Last edited: Feb 18, 2008
3. Feb 18, 2008

jj8890

oops...i mean 4.1 would be x and 0 would be y...

Last edited: Feb 18, 2008
4. Feb 18, 2008

jj8890

is that right?

5. Feb 18, 2008

jj8890

I try to multiply these two together: 1.5xi+1.7yj and 4.1xi+0.0yj and I get 6.15 but this is not right. What am i doing wrong?

6. Feb 18, 2008

physixguru

it shall save a lot of trouble if ya try integration dear..

>>Work Done = integral of vector F .dx ; integrate x from 0 to 4.1.

this shall simplify the problem much...

7. Feb 18, 2008

belliott4488

I think I see what's going on. jj8890 is assuming that the force is constant, as I was at first. Since the force is given in units of N/m, then I see that the x and y appearing in the force definition are not subscripts, but are variables of displacement. We must have a spring here.

So, jj8890 - try what physixguru suggested - do your dot product, but do it for each point along the motion in the x-direction, i.e. integrate. (If you know how to use Hook's Law for springs, you could jump right to the answer, but doing the integral essentially derives the expression for the potential energy of a spring, which goes with Hook's Law.)