# Sled and a rope

1. Oct 20, 2006

### kbyws37

A sled is dragged along a horizontal path at a constant speed of 1.5 m/s by a rope that is inclined at an angle of 30.0° with respect to the horizontal (see the figure below). The total weight of the sled is 470 N. The tension in the rope is 250 N. How much work is done by the rope on the sled in a time interval of 5.00 s?

I'm not sure if i'm on the right track.
I'm using the equation
W=K+U =E
E = (mv^2)/2 + mgy
but i don't know where I would put time and where

250cos(30) would go.

or maybe i'm just totally off.

2. Oct 20, 2006

### Hootenanny

Staff Emeritus
Think about the definition of work done (in terms of force).

3. Oct 20, 2006

You know the velocity, and you know the time. So, you know the displacement, too. I'm sure you'll know how to carry on from this point.

4. Oct 20, 2006

### kbyws37

I used x = ((Vi + Vf)t)/2 and found the displacement to be 3.75
Then I used the equation:

E = (mv^2)/2 + mgx = W
= ((470)(1.5^2))/2 + ((470)(9.8)(3.75))
= a huge number (which is not correct)

If i use the equation
W = F*x*cos()
W = 470*3.75*cos(30)
W = 1526.37
which says that I am wrong

5. Oct 20, 2006

Ever heard of the simple expression for constant velocity $$x = v\cdot t$$?