Recent content by foxat

Thanks! I have approached this problem by using total F = ma, which equals the force exerted by the locomotive on the caboose minus the frictional force. So ma = F of the locomotive - 0.51, and thus F of the locomotive = ma + 0.51. The answer i came up with is (2.1 + 1.7)(3.4) + 0.51, or 13.43...

A 2.1 kg toy locomotive is pulling a 1.7 kg caboose. The frictional force of the track on the caboose is 0.51 N backward along the track. If the train is accelerating forward at 3.4 m/s2, what is the magnitude of the force exerted by the locomotive on the caboose?