# Skater pulling another skater with a rope, f=ma

1. Mar 31, 2013

### VidsEpic

1. The problem statement, all variables and given/known data

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

2. Relevant equations

F = ma

3. 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?

2. Mar 31, 2013

### 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.

3. Mar 31, 2013

### Staff: Mentor

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

What book is it?

4. Mar 31, 2013

### Staff: Mentor

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

5. Mar 31, 2013

### VidsEpic

So that was it!

Thanks a lot!

6. Mar 31, 2013

### VidsEpic

Last edited by a moderator: May 6, 2017
7. Mar 31, 2013

### 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.

8. Mar 31, 2013

### Staff: Mentor

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.

9. Jul 11, 2017

### strugglingstudent

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

10. Jul 11, 2017

### Staff: Mentor

Look again. The force is given.