# Work and Finding Power

1. Oct 7, 2009

### danest

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

A force = (5.10 N) + (9.50 N) + (7.50 N) acts on a 2.30 kg mobile object that moves from an initial position of i = (1.30 m) - (9.70 m) + (7.90 m) to a final position of f = -(1.90 m) + (4.80 m) + (8.00 m) in 4.50 s. Find (a) the work done on the object by the force in the 4.50 s interval, (b) the average power due to the force during that interval, and (c) the angle between vectors i and f.

2. Relevant equations

W= F dot d

W= F*d

3. The attempt at a solution

First I attempted to solve for a by doing the dot product of the force vector and the d final but that didn't work. Then I found the magnitudes of F and d final and multiplied them but then that didn't work either. Am I supposed to use F= ma and find the acceleration? but if its that what am I supposed to find after?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Solved

Last edited: Oct 7, 2009
2. Oct 7, 2009

### rl.bhat

You better write the problem in the proper vector form. Like
F = (5.10 N)i + (9.50 N)j + (7.50 N)k
Initial position vector A = (1.30 m)i - (9.70 m)j + (7.90 m)k
Final position vector B = -(1.90 m)i + (4.80 m)j + (8.00 m)k
The displacement vector d = (B - A)
Now find w = F.d