Dynamics how to calculate maximum height on incline

In summary, a 5.0kg block initially traveling at 11m/s moves up a 30 degree incline with a frictional force of 9.4N acting against it. Using the equations vf^2=vi^2+2ad and Fnet=ma, the maximum vertical height, h, is found to be 4.5m. The direction of the x component of the weight and the friction force play a key role in determining the acceleration and final position of the block on the incline.
  • #1
jakeginobi

Homework Statement


A 5.0kg block initally traveling 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,

Homework Equations


vf^2=vi^2+2ad, Fnet=ma

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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  • #2
What direction is your x component of the weight?
 
  • #3
TomHart said:
What direction is your x component of the weight?
http://imgur.com/a/ndFug here's the picture of the diagram
 
  • #4
jakeginobi said:
here's the picture of the diagram
I was hoping for a free body diagram.
 
  • #5
Anyway, just check your signs.
 
  • #7
jakeginobi said:
[(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?
 
  • #8
I thought friction always acts opposite to where the object is moving?
 
  • #9
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.
 
  • #10
TomHart said:
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
 
  • #11
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.
 
  • #12
TomHart said:
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?
 
  • #13
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.
 
  • #14
TomHart said:
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
 
  • #15
PhysicsForum block diagram 1.jpg
 
  • #16

FAQ: Dynamics how to calculate maximum height on incline

What is the formula for calculating maximum height on an incline?

The formula for calculating maximum height on an incline is h = (v^2*sin^2(theta)) / 2g, where h is the maximum height, v is the initial velocity, theta is the angle of the incline, and g is the acceleration due to gravity.

How does the angle of the incline affect the maximum height?

The angle of the incline directly affects the maximum height. As the angle increases, the maximum height decreases. This is because more of the initial velocity is directed towards the motion along the incline, rather than straight up.

What is the significance of the initial velocity in calculating maximum height?

The initial velocity is crucial in calculating maximum height because it determines the amount of energy that the object has at the beginning of its motion. This energy is then used to overcome the force of gravity and reach the maximum height.

Can the formula for calculating maximum height on an incline be applied to all objects?

The formula for calculating maximum height on an incline is applicable to objects that follow a projectile motion, such as a ball rolling down an incline or a car driving up a ramp. It may not be accurate for objects with complex motion.

Is there a maximum possible height that an object can reach on an incline?

There is no specific maximum height that an object can reach on an incline. The height will vary depending on the initial conditions, such as the angle and initial velocity, as well as external factors like friction. However, the maximum height will always be less than the height of the incline itself.

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