Calculating Force Exerted by 1000kg Car on 450kg Trailer

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A 1000kg car exerts a horizontal force of 3500N to pull a 450kg trailer, and the frictional force for the trailer, calculated using a coefficient of 0.15, must be considered in determining the force exerted on the trailer. The frictional force is derived from the equation (0.15)(450)(9.81), which needs to be either added to or subtracted from the car's force. The discussion also shifts to a separate problem involving a car's skid mark of 88m on a rainy day with a friction coefficient of 0.42, prompting a calculation of the car's speed at the moment brakes were applied. It is noted that the car's mass is irrelevant in this context due to the cancellation of mass in the equations used. Understanding these principles is crucial for solving both problems effectively.
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The problem is, a 1000kg car pulls a 450kg trailer. The car exerts a horizontal force of 3500N against the ground in order to accelerate. What force does the car exert on the trailer? Assume a friction coefficient of 0.15 for the trailer.
 
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dabouncerx24 said:
The problem is, a 1000kg car pulls a 450kg trailer. The car exerts a horizontal force of 3500N against the ground in order to accelerate. What force does the car exert on the trailer? Assume a friction coefficient of 0.15 for the trailer.

You need to either add the frictional force to the 3500N or subtract it. The frictional force equals (0.15)(mg) or (0.15)(450)(9.81).
 
I don't get why adding the friction force to 3500N is logical? Can you please explain it to me.
 
Do the forces analysis...
 
Well I figured out that problem.

Here is another one, please assist me.

At an accident scene on a level road, investigators measure a car;s skid mark to be 88m long. It was a rainy day and the coefficient of friction was estimated to be 0.42. Use these dta to determine the speed of the car when the driver slammed on the brakes (why does the car's mass not matter).

So far I figured I would use Fd=KE-KE' so all the ms will cancel each other.
 
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