annoying physics question

[mustymullet]

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Hello, i know this probably should go in the mcat study questions but could you help me out? The question is:

A coffee cup on the dashboard of a car slides forward on the dash when the driver decelerates from 38.7 km/hr to rest in 3.69s or less, but not if he decelerates in a longer time. What is the coefficient of static friction between the cup and the dash?

I already calculated the acceleration (or deceleration for that matter) to be
-2.91m/s^2. I just dont know how to get the coefficient if they dont give you the force of friction or the Fn (normal force).

 

[cyclingmo]

Hello, i know this probably should go in the mcat study questions but could you help me out? The question is:

A coffee cup on the dashboard of a car slides forward on the dash when the driver decelerates from 38.7 km/hr to rest in 3.69s or less, but not if he decelerates in a longer time. What is the coefficient of static friction between the cup and the dash?

I already calculated the acceleration (or deceleration for that matter) to be
-2.91m/s^2. I just dont know how to get the coefficient if they dont give you the force of friction or the Fn (normal force).

The clue is that the cup does not slide if the decelaration takes any longer (i.e. is less). So the 2.91 m/s2 decelaration that you have is just when the friction force gives out.

I'm rusty, but it might go something like this: F=ma and in this case F=umg where u= coeff of static friction.

So you'll have umg=ma, the m's cancel out and that gives you u=a/g which will be 2.91/9.81 or 0.297 for friction.

Am I right or wrong? I'm afraid momentum might be involved, but that conservative, so maybe not.

 

[mustymullet]

The clue is that the cup does not slide if the decelaration takes any longer (i.e. is less). So the 2.91 m/s2 decelaration that you have is just when the friction force gives out.

I'm rusty, but it might go something like this: F=ma and in this case F=umg where u= coeff of static friction.

So you'll have umg=ma, the m's cancel out and that gives you u=a/g which will be 2.91/9.81 or 0.297 for friction.

Am I right or wrong? I'm afraid momentum might be involved, but that conservative, so maybe not.

THANK YOU IT WORKED!!🙂 YOU ARE VERY SMART

 

[cyclingmo]

I'm glad it worked, just make sure you also understand the concept. Physics is easy once you master the concepts.

 

[adoggie]

Sorry this might sound stupid, but how did you find the acceleration? Which kinematic formula did you use?

 

[sehnsucht]

Sorry this might sound stupid, but how did you find the acceleration? Which kinematic formula did you use?

since 38.7 km/h = 10.75 m/s, you use the following values in the forumula: Vf=Vo + a*t. Solve for a by dividing 10.75 m/s/ 3.69s = 2.91m/s. The negative sign is a direction convention depending on which side - left or right - you designate as positive.

 

[adoggie]

so Vf = 10.75, and Vo = 0?

 

[sehnsucht]

so Vf = 10.75, and Vo = 0?

In this case, Vf would be 0 since you are decelerating from motion, right? So, Vf=0 and Vo=10.75m/s.. Plug and chug to solve for "a."

 

[adoggie]

Now I get it! Thanks.

 

[osli]

Always look at questions like this from a "big picture, intuitive perspective" first. Then knowing what to do with the number given and equations in your head will be simple.

Here's something to always remember about these types of problems... stuff sliding due to an acceleration and being asked about friction. It doesn't matter how much the coffee cup weighs. If you increase the weight, the force of friction is greater, but so is the force developed by acceleration trying to move it. They exactly cancel. Also, the size of the cup doesn't matter. Increasing surface area is offset by decreased force per square area. Total friction force developed is unaffected.