Thread: force problem

Jane and John, with masses if 50 kg and 60 kg, respectively, stand on a frictionless surface 10 m apart. John pulls on a rope that connects him to Jane, giving Jane an acceleration of 0.92 m/s2. At what rate will John accelerate? What is the direction of his acceleration?

If I understand the problem correctly then you should use the concept of net force. Jane's net force is (.92 m/s^2) * 60kg. Now use the formula F_net=ma, and solve for John's acceleration.

Originally Posted by Jameson

If I understand the problem correctly then you should use the concept of net force. Jane's net force is (.92 m/s^2) * 60kg. Now use the formula F_net=ma, and solve for John's acceleration.

I'll add a bit of detail here. Jane is exerting a force (F) on John. By Newton's third law, the force that John exerts on Jane will also be F, in the opposite direction. This is why you can do as Jameson suggests.

-Dan

Yeah, if I remember correctly Newton's Third Law states that to every action there is an equal and opposite reaction.
Exactly what topsquark said....