# Calculate work done if a 25N force is acting in a direction of vector (2,3,-1)?

• Feb 7th 2009, 03:49 AM
Random-Hero-
Calculate work done if a 25N force is acting in a direction of vector (2,3,-1)?
Calculate the work done if a 25N force acting in a direction of vector (2,3,-1) moves an object from P(2,-3,1) to Q(5,0,2).

Anyone know how I'd go about solving this?

All I knows is I gotta use 'Work = Force (dot) Displacement' at some points :|
• Feb 7th 2009, 04:39 AM
HallsofIvy
Quote:

Originally Posted by Random-Hero-
Calculate the work done if a 25N force acting in a direction of vector (2,3,-1) moves an object from P(2,-3,1) to Q(5,0,2).

Anyone know how I'd go about solving this?

All I knows is I gotta use 'Work = Force (dot) Displacement' at some points :|

Yes, and the only thing you need to do is find the Force and Displacements vectors. The displacement vector is easy: the vector from (2, -3, 1) to (5, 0, 2) is $(5- 2)\vec{i}+ (0-(-3))\vec{j}+ (2- 1)\vec{k}= 3\vec{i}+ 3\vec{j}+ vec{k}$ which we can also write as <3, 3, 1> (I prefer to use < > for vectors rather than ( ) because that is too easy to confuse with points.)

A little harder is the force vector. In order to use the "direction of vector <2, 3, -1>" we need to remove the length. We can do that by dividing by its length which is $\sqrt{2^2+ 3^2+ (-1)^2}= \sqrt{4+9+1}= \sqrt{14}$. So $\frac{1}{\sqrt{14}}<2, 3, -1>$ is a unit vector in that direction and since the force has strength 25N, the force vector is $\frac{25}{\sqrt{14}}<2, 3, -1>$.

Find the dot product of <3, 3, 1> and $\frac{25}{\sqrt{14}}<2, 3, -1>$
• Feb 7th 2009, 06:14 AM
Random-Hero-
<3, 3, 1> and http://www.mathhelpforum.com/math-he...bab7494f-1.gif

= <3,3,1> dot <50/root14, 75/root14, -25/root14>

= 150/root14 + 225/root14 -25/root14

= 350/root14

So the value of 350/root14 would be the amount of work done in Joules?