Translational Motion and Center of Mass

[The_Fritz]

Homework Statement

A 280-kg flatcar 25m long is moving with a speed of 6.0m/s along horizontal frictionless rails. A 95-kg worker starts walking from one end of the car to the other in the direction of motion, with speed 2.0m/s with respect to the car. In the time it takes for him to reach the other end,how far has the flatcar moved?

Homework Equations

Falls under the category of Center of Mass and Translational Motion
Ma_cm = [tex]\Sigma[/tex]F_ext

The Attempt at a Solution

Since the rails are frictionless I figured the mass of both the flatcar and the worker are irrelevant.
Time it takes worker to walk to end of flatcar = (25m)(second/2.0m) = 12.5 s
Distance flatcar travels in the time it takes worker to walk to the other end =
(6.0m/s)(12.5s)=75m
Is this the correct answer or am I missing something? Input will be very much appreciated.

 

[anathema]

Hello! Your solution is correct. The key equation you used, Ma_cm = \SigmaF_ext, relates the mass of an object to its acceleration when all external forces acting on it are considered. In this case, since there are no external forces acting on the flatcar and worker system, the mass of the flatcar and worker are indeed irrelevant. The solution you provided correctly considers the time it takes for the worker to walk to the other end of the flatcar and the distance the flatcar travels in that time. Keep up the good work!

 

[kerimek]

Your approach is correct. The distance traveled by the flatcar is equal to the distance the worker walks in the time it takes him to reach the other end, since both are moving at the same speed in the same direction. Therefore, the flatcar moves 75m while the worker walks 25m. This assumes that the worker's speed relative to the car remains constant throughout the walk.