# 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

### radou

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

### radou

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

6. Oct 21, 2006

### Hootenanny

Staff Emeritus
Although radou's equation is simpler to use your equation above is valid. If you substitute in the the correct values you should obtain the correct answer. However, 3,75 is not correct. The rest of your working;
Looks good except for the erroneous value for displacement.

Last edited: Oct 21, 2006
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