• jmartinez638
Will medal and fan: A 45 kg block slides down an incline that is angled at 41 degrees from horizontal. If the coefficient of kinetic friction between the block and the inclined surface is μk = 0.45, what is the block’s acceleration?
  • Stacey Warren - Expert
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  • jamiebookeater
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  • anonymous
Here is our scenario:
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  • anonymous
Wow, that's a lot we have here! Let's break it down a bit. What are we looking for here? We are asked to find the acceleration down the inclined plane. We need to create a modeled equation in order to do so. We can do this by using Newton's 2nd Law!\[\huge \sum \text{F}=\text{F}_g-\text{F}_f=ma\]Next we need to find out what exactly Fg and Ff are. Let's start with Fg. Fg is the component of the gravitational force that pushes the box down the inclined plane. By using trigonometry, we can find that this Fg is:\[\huge \text{F}_g= mg \sin(\theta)\]Now let's move onto Ff In a general case, we can easily define Ff as:\[\huge \text{F}_f=\mu \text{F}_\text{normal}\]Well now we have to find the normal force! Ugh! By looking at our diagram that we created, we can use trigonometry to see that for an inclined plane,\[\huge \text{F}_\text{normal}= mg \cos(\theta)\]So now that we have the normal force, that means that our frictional force is:\[\huge \text{F}_f=\mu mg \cos(\theta)\] Almost done! Now we just have to plug this into our modeled equation that we first stated! \[\huge mg \sin(\theta)-\mu mg \cos(\theta)=ma\]From there we can see that by solving for a, all of our masses will cancel out! Wait a minute.. that means that the acceleration solely depends on gravity and the angle of the incline!

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