Newton's 2nd Law Question PHYCS110

[NanoTech]

Hey everyone~

A 2.0 kg toy locomotive is pulling a 1.0 kg caboose. The frictional force of the track on the caboose is 0.50 N backward along the track. Of the train is accelerating foward at 3.0 m/s^2, what is the magnitude of the force exerted by the locomotive on the caboose?

So I'm guessing that I use the equation: F = ma, but I'm not sure how to get started. The answer in the back of the book is: 3.5 N. Thanks for your input~David.W

 

[AKG]

Hey everyone~

A 2.0 kg toy locomotive is pulling a 1.0 kg caboose. The frictional force of the track on the caboose is 0.50 N backward along the track. Of the train is accelerating foward at 3.0 m/s^2, what is the magnitude of the force exerted by the locomotive on the caboose?

So I'm guessing that I use the equation: F = ma, but I'm not sure how to get started. The answer in the back of the book is: 3.5 N. Thanks for your input~David.W

$$F_{net}\ =\ ma$$

Now, you know the mass. Also, you know the acceleration of the train, so it should be pretty obvious as to what the acceleration of the caboose is. (Think about it, if my body is accelerating at X, then what would be the acceleration of my head, arm, finger... ?) So, you should be able to find the net force on the train. Since the train is accelerating parallel to the ground, find the net forces acting on the caboose in the direction parallel to the ground. What are these forces? Figure them out, and set up an equation to find the force exerted by the engine on the caboose.

 

[M98Ranger]

Hi David,

You're on the right track! To find the magnitude of the force exerted by the locomotive on the caboose, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma).

In this scenario, the net force acting on the caboose is the force exerted by the locomotive (forward) minus the frictional force of the track (backward). So we can set up the equation as:

F - 0.50 N = (1.0 kg)(3.0 m/s^2)

We can rearrange this equation to solve for F:

F = (1.0 kg)(3.0 m/s^2) + 0.50 N

F = 3.0 N + 0.50 N

F = 3.50 N

So the magnitude of the force exerted by the locomotive on the caboose is 3.50 N, which is consistent with the answer in the back of the book.

Hope this helps! Let me know if you have any other questions or need clarification. Happy studying!