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Dynamics how to calculate maximum height on incline

  • Thread starter jakeginobi
  • Start date
1. Homework Statement
A 5.0kg block initally travelling at 11m/s moves up a 30degrees incline. A frictional force of 9.4N acts on the block as it moves up the incline. What maximum vertical height, h, will the block reach?
m = 5.0kg, vi = 11m/s, vf = 0m/s, Fr = 9.4N,
2. Homework Equations
vf^2=vi^2+2ad, Fnet=ma

3. The Attempt at a Solution
I tried finding the acceleration by doing F|| - Fr = ma [(5.0kg)(9.8)(sin30)-9.4N=(5.0kg)a]
when I found the acceleration, I used the equation vf^2=vi^2+2ad to find d, by assuming vf = 0m/s, and vi = 11m/s, which d would be = 20.033 m. After that I did 20.033m x sin30 which is 10m. The answer is supposed to be 4.5m
 
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What direction is your x component of the weight?
 
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Anyway, just check your signs.
 
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[(5.0kg)(9.8)(sin30)-9.4N=(5.0kg)a]
According to the free body diagram that you just posted, isn't the x component of the weight in the same direction as the friction force?
 
I thought friction always acts opposite to where the object is moving?
 
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Oh, I just now noticed the direction of your x component of the weight. You have it pointing in the positive x direction, but the weight vector is obviously to the left of the y axis. So the x component of the weight should be pointing in the negative x direction.
 
Oh, I just now noticed the direction of your x component of the weight. You have it pointing in the positive x direction, but the weight vector is obviously to the left of the y axis. So the x component of the weight should be pointing in the negative x direction.
http://imgur.com/a/ZRYhf so like this? I kinda understand it but I'm just confused about the Fnet. I thought the F net would be the applied force minus the friction instead of adding the frictional force since it is accelerating up the ramp
 
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It's like a car coasting up a hill. (Assume there is no friction.) The car does not have to apply the brakes to decrease the car's speed because the "x component" of the car's weight is in the opposite direction of the car's velocity. However, if the car applies the brakes also, the car will slow down more quickly because both of those forces are acting in the same direction - that is, in the opposite direction of the velocity. That is the exact situation you have in this problem.
 
It's like a car coasting up a hill. (Assume there is no friction.) The car does not have to apply the brakes to decrease the car's speed because the "x component" of the car's weight is in the opposite direction of the car's velocity. However, if the car applies the brakes also, the car will slow down more quickly because both of those forces are acting in the same direction - that is, in the opposite direction of the velocity. That is the exact situation you have in this problem.
oh, what if it's going down then you would minus the frictional force right?
 
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If the velocity of the block is down the slope (the negative x direction), in that case the friction force would be up the slope (the positive x direction) in the opposite direction of the velocity. But the x component of the weight would still be in the negative x direction. So, when the block is sliding up the slope, both forces are in the same direction (negative x direction), but when sliding down the slope, one force is in the negative x direction and one force is in the positive x direction.
 
If the velocity of the block is down the slope (the negative x direction), in that case the friction force would be up the slope (the positive x direction) in the opposite direction of the velocity. But the x component of the weight would still be in the negative x direction. So, when the block is sliding up the slope, both forces are in the same direction (negative x direction), but when sliding down the slope, one force is in the negative x direction and one force is in the positive x direction.
oh alright thanks :D, if possible could you sketch a free body diagram sorry to bother you :/ I think a degree would be easier for me to understand
 
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PhysicsForum block diagram 1.jpg
 

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