How Much Force Does a Toy Locomotive Exert on a Caboose?

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A 2.1 kg toy locomotive is pulling a 1.7 kg caboose with a frictional force of 0.51 N acting backward. The train accelerates forward at 3.4 m/s², prompting a calculation of the force exerted by the locomotive on the caboose. The initial approach used the equation F = ma, incorporating the frictional force, resulting in a calculated force of 13.43 N. However, clarification is sought regarding the net force required for the caboose's acceleration and the use of Free Body diagrams. Understanding these concepts is essential for accurately determining the locomotive's force on the caboose.
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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?
 
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foxat said:
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?
Hello foxat. Welcome to PF !

What have you tried?

Where are you stuck?

You must show an attempt, before we can help you. --- those are the rules here at PF.
 
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 N. Am I approaching this the right way?
 
foxat said:
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 N. Am I approaching this the right way?
No.

What net force is needed to accelerate the caboose at 3.4 m/s2 ?

Also, do you know about Free Body diagrams ?
 
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