Solve Work & Angle Homework: Plane Towing Glider & Water Skier

[songoku]

Homework Statement

1. A small plane tows a glider at constant speed and altitude. If the plane does 2.00 x 10⁵ J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the tow rope and the horizontal?

2. Water skiers often ride to one side of the center line of a boat as shown below. In this case, the ski boat is traveling at 15 m/s and the tension in the rope is 75 N. If the boat does 3500 J of work on the skier in 50.0 m, what is the angle θ between the tow rope and the center line of the boat?

Homework Equations

W = F d cos θ
W = Δ KE

The Attempt at a Solution

1.
W = F d cos θ
2 x 10⁵ = 2560 x 145 cos θ
θ = 57.4^o
I'm not sure because I don't know the picture of plane tows a glider. I am only using the formula

2.
W = F d cos θ
3500 = 75 x 50 cos θ
θ = 21.04^o
Is the speed given in the question not important?

Thanks

 

[TSny]

All looks good to me. The speed is irrelevant in part b.

 

[CWatters]

and me. Speed irrelevant in both parts as the problem is the same.

Speed information would have be needed had the question given you "power" rather than "work".

 

[songoku]

All looks good to me. The speed is irrelevant in part b.

and me. Speed irrelevant in both parts as the problem is the same.

Speed information would have be needed had the question given you "power" rather than "work".

Thanks

 

[Nickie]

for your question! I would like to clarify some concepts and provide a more detailed response to the content.

First, let's define some terms and equations that are relevant to this problem. Work is the measure of energy transferred when a force is applied over a distance. It is represented by the letter W and is measured in joules (J). The equation for work is W = Fd, where F is the force applied and d is the distance over which the force is applied. In this problem, the work is done by the force of the plane or boat pulling the glider or skier respectively.

The angle between the tow rope (or the center line of the boat) and the horizontal is represented by θ. This angle is important because it affects the amount of work done on the glider or skier. The equation for work with an angle is W = Fd cos θ. This equation takes into account the angle between the force and the direction of motion.

Now let's look at the solutions to the two problems.

1. In the first problem, we are given the work done (2.00 x 105 J) and the distance over which the force is applied (145 m). The equation for work is W = Fd, so we can rearrange the equation to solve for the force: F = W/d. Plugging in the values, we get F = (2.00 x 105 J)/(145 m) = 1379 N. However, this is not the tension in the tow rope. The tension is the force that the rope exerts on the glider, which is equal in magnitude but opposite in direction to the force applied by the plane. So the tension is actually 1379 N in the opposite direction.

To find the angle θ, we can use the equation W = Fd cos θ and solve for cos θ. Plugging in the values, we get cos θ = (2.00 x 105 J)/(1379 N x 145 m) = 0.8309. To solve for θ, we take the inverse cosine (cos^-1) of both sides, giving us θ = 33.4o. This is the angle between the tow rope and the horizontal.

2. In the second problem, we are given the work done (3500 J) and the distance over which the force is applied (50.