# Calculating Work Done by a Rope on a Sled

• kbyws37
In summary, a sled is being dragged at a constant speed of 1.5 m/s by a rope inclined at 30.0° with a weight of 470 N and a tension of 250 N. To find the work done by the rope in 5.00 s, one can use the equation W = F*x*cos(), where the displacement can be calculated using x = v*t. However, in the given conversation, the value for displacement was incorrect.
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.

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

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.

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

kbyws37 said:
I used x = ((Vi + Vf)t)/2 and found the displacement to be 3.75

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

kbyws37 said:
I used x = ((Vi + Vf)t)/2 and found the displacement to be 3.75
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;
kbyws37 said:
If i use the equation
W = F*x*cos()
Looks good except for the erroneous value for displacement.

Last edited:

## 1. How do you calculate the work done by a rope on a sled?

The work done by a rope on a sled can be calculated by multiplying the force applied by the rope by the distance the sled moves in the direction of the force. This can be expressed as W = Fd, where W is work, F is force, and d is distance.

## 2. What units are used to measure work?

The SI unit for work is joule (J), which is equivalent to a kilogram-meter squared per second squared (kg*m^2/s^2). Other common units for work include calories and foot-pounds.

## 3. Can work be negative?

Yes, work can be negative. This occurs when the applied force is in the opposite direction of the displacement of the object. In the case of a rope pulling a sled, if the sled moves in the opposite direction of the force applied by the rope, the work done by the rope will be negative.

## 4. What factors affect the work done by a rope on a sled?

The work done by a rope on a sled is affected by the magnitude of the applied force, the distance the sled moves, and the angle between the force and the displacement. The work done will be greater if the force is larger, the sled moves a greater distance, or if the force and displacement are in the same direction.

## 5. How is the work-energy theorem related to calculating work done by a rope on a sled?

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In the case of a sled being pulled by a rope, the work done by the rope will result in an increase in the sled's kinetic energy, which can be calculated using the formula KE = 1/2mv^2, where m is mass and v is velocity.

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