Skater pulling another skater with a rope, F=ma

[VidsEpic]

Homework Statement

Two people, each with a mass of 70 kg, are wearing inline
skates and are holding opposite ends of a 15 m rope. One
person pulls forward on the rope by moving hand over hand
and gradually reeling in more of the rope. In doing so, he
exerts a force of 35 N [backwards] on the rope. This causes
him to accelerate toward the other person. Assuming that
the friction acting on the skaters is negligible, how long will
it take for them to meet? Explain your reasoning. T/I

Homework Equations

F = ma

The Attempt at a Solution

35N / 70 kg = 0.5 m/s/s

displacement = 0.5*a*t^2
7.5 = 0.25 * t^2

Since they are both accelerating towards each other because of Newton's third law, only half the distance is required to be traveled, so it takes 5.47 seconds for them to meet.

The answer at the answer key of my book says that it takes 7.7 seconds, they have used 15m in the equation instead of 7.5. If tension is the same at both ends of the rope, why would the person need to travel the whole 15m? the other skater can't be stationary?

 

[HallsofIvy]

You are correct about the 7.5 meters. Your error is in calculating the acceleration. Since the person pulling on the rope is accelerating as well as the person being accelerated, the 35 N force is accelerating 140 kg, not 70 kg.

 

[Doc Al]

Since they are both accelerating towards each other because of Newton's third law, only half the distance is required to be traveled, so it takes 5.47 seconds for them to meet.

The answer at the answer key of my book says that it takes 7.7 seconds, they have used 15m in the equation instead of 7.5. If tension is the same at both ends of the rope, why would the person need to travel the whole 15m? the other skater can't be stationary?

I'd say that you are correct and the book is wrong.

What book is it?

 

[Doc Al]

You are correct about the 7.5 meters. Your error is in calculating the acceleration. Since the person pulling on the rope is accelerating as well as the person being accelerated, the 35 N force is accelerating 140 kg, not 70 kg.

35 N is the tension in the rope, which acts on each of them separately. The 35 N force accelerates each 70 kg mass.

 

[VidsEpic]

So that was it!

Thanks a lot!

 

[VidsEpic]

I'd say that you are correct and the book is wrong.

What book is it?

http://www.lakeheadschools.ca/scvi_staff/childs/Gr11_physics_web/downloadable_content/unit3/textpdf3/phys11_3_5.pdf

here it is:

Nelson Physics 11 university preparation

http://www.torontopubliclibrary.ca/detail.jsp?Entt=RDM2821349&R=2821349

 

[VidsEpic]

If you look at the other questions, you will find some very vague and strange questions. I encounter a lot of irregular patterns. For example sometimes they ask for a force, and I am not sure whether they want the net force, or the applied or the tension, and in the answer key, I actually figure out what they were looking for. So the book could possibly be flawed.

 

[Doc Al]

Unfortunately, I'm not familiar with that book.

I don't see anything unusual about the question you brought up in this thread, but it's not uncommon for the answer key to have mistakes. Feel free to bring up other problems, if they seem a bit irregular.

 

[strugglingstudent]

How can this problem be done when a force is not given?

 

[Doc Al]

How can this problem be done when a force is not given?

Look again. The force is given.

 

[iiSummerboy21]

I'd say that you are correct and the book is wrong.

What book is it?

I got 7.7 so I did the same steps as him
7.5 = 0.5*0.25*t^2
7.5/0.125 = t^2
60 = t^2
7.7 = t

 

[kuruman]

Welcome to PF @iiSummerboy21.

You realize that this thread is more than 9 years old. That said, your method is correct but the acceleration is not 0.25 m/s² as implied in your equation
7.5 = 0.5*0.25*t^2
The force of tension acting on one person is T = 35 N. If the person's mass is m = 70 kg, the person's acceleration is
a = T/m = 35/70 =1/2 m/s².

 

[SammyS]

I got 7.7 so I did the same steps as him
7.5 = 0.5*0.25*t^2
7.5/0.125 = t^2
60 = t^2
7.7 = t

@iiSummerboy21 ,

This thread was last replied to more than 5 years ago and is more than 9 years old.

That being said:

I assume you plugged into the expression $\dfrac 1 2 a t^2$ .

How did you get 0.25 (m/s²) for acceleration?

(@kuruman beat me to it !)