How Is Net Work Calculated When Pulling a Boat?

[mikefitz]

http://img517.imageshack.us/img517/6533/c6p4ty2.jpg

The drawing shows a boat being pulled by two locomotives through a canal of length 1.04 km. The tension in each cable is 4.27 x 103 N, and theta = 20.0o. what is the net work done on the boat by the two locomotives?


Here is what I've done so far:

Force=4.27*10^3
delta(r)=1.04
theta=20

W=2* [4.27*10^3 (1.04) (cos20) ]

=8345J

What am I doing wrong here? thanks

 

[Doc Al]

The distance is given in km--convert to meters.

 

[kyle8921]

Your calculation is correct, but it is not the net work done on the boat by the two locomotives. In order to find the net work, we need to consider the direction of the force and displacement.

The net work is the sum of the work done by both locomotives, which can be calculated using the formula W = Fdcosθ, where F is the force, d is the displacement, and θ is the angle between the force and displacement.

In this case, the force and displacement are both in the same direction, so θ = 0. Therefore, the net work done by both locomotives can be calculated as:

W = (4.27*10^3 N)(1.04 km)(cos 0°) + (4.27*10^3 N)(1.04 km)(cos 0°)

= 8,832 J

This is the total amount of work done by both locomotives on the boat.