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

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Homework Help Overview

The problem involves a toy locomotive pulling a caboose, with specific weights and a frictional force acting on the caboose. The scenario includes calculating the force exerted by the locomotive while the system is accelerating.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law, with one attempting to express the force exerted by the locomotive in relation to the total mass and frictional force. Questions arise regarding the net force required for acceleration and the use of free body diagrams.

Discussion Status

Some participants are exploring different interpretations of the problem and questioning the assumptions made in the calculations. Guidance has been offered regarding the need to consider the net force for the caboose's acceleration.

Contextual Notes

Participants are reminded of the forum rules requiring them to show attempts before receiving assistance, which influences the nature of the discussion.

foxat
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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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