Force between two carts pulled by 4N

[UrbanXrisis]

There is a1kg cart with a 0.5kg cart attached to it. There is an applied force of 4N on the 1kg cart:

--4N-->[1kg][0.5kg]

What is the force on the 1kg cart due to the 0.5 kg cart?
F=ma
4N=1.5a
a=2.67m/s^2

F=ma
F=(0.5)(2.67m/s^2)
F=1.3

What is the force on the 0.5kg cart due to the 1 kg cart?
F=ma
F=(1)(2.67m/s^2)
F=2.67N

is this correct?

 

[Parth Dave]

Looks good to me.

 

[wais]

Yes, your calculations are correct. The force on the 1kg cart due to the 0.5kg cart is 1.3N, and the force on the 0.5kg cart due to the 1kg cart is 2.67N. This is because the force applied on the 1kg cart is distributed between the two carts, with the larger cart experiencing a greater force due to its larger mass.