How Far Will the Skateboarder Travel Up the Second Incline?

In summary: I don't know, that's what the example in the books says.g is always a positive constant. It doesn't mean that acceleration has to be positive though. It depends on how you define positive or negative direction. You get to define this, by the way. For example you could define up as being the positive direction for a problem involving a free-falling object. In that case, since the object is accelerating down, a = -g. But note that g is still positive even though the acceleration isn't. Sometimes you'll need to stick a negative sign in front of g to make the acceleration negative, but that doesn't make g itself is negative.
  • #1
phizics09
38
1

Homework Statement


A skateboarder slides down a frictionless ramp inclined at an
angle of 30° to the horizontal. He then slides across a frictionless
horizontal floor and begins to slide up a second incline at an angle
of 25° to the horizontal. The skateboarder starts at a distance of
10 m from the bottom of the first incline.How far up the second
incline will he go if the coefficient of kinetic friction on the second
incline is 0.10?

Homework Equations


Fnet=ma
a=gsin(theta)

The Attempt at a Solution


So I got the first part right-- I found V2 to be 9.9 m/s. So for the second ramp, I used the formula Fnet=F(parallel)-u(Fn).
ma=mgsin25 deg. -0.1(cos 25 deg.)
I got an acceleration of 0.33 m/s^2, but the answer at the back is different. Also, I don't know whether g should be positive or negative.

Thanks in advance!
 
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  • #2
phizics09 said:
So I got the first part right-- I found V2 to be 9.9 m/s. So for the second ramp, I used the formula Fnet=F(parallel)-u(Fn).
ma=mgsin25 deg. -0.1(cos 25 deg.)
I got an acceleration of 0.33 m/s^2, but the answer at the back is different. Also, I don't know whether g should be positive or negative.

Thanks in advance!
The net force is always in the direction of the acceleration. On the 2nd incline, which way does the gravity component act, down or up the incline? Which way does friction act, down or up the incline? Which way does acceleration act? be consistent with your choice of plus and minus signs.
 
  • #3
phizics09 said:
So for the second ramp, I used the formula Fnet=F(parallel)-u(Fn).
I assume that "F(parallel)" is the component of the gravitational force that is parallel to the ramp? That's fine if it is, I just want to make sure that we're both talking about the same thing.

For this particular problem, is "F(parallel)" in the same direction as the frictional force or in the opposite direction? Draw a free body diagram if that helps.

[Edit: PhanthomJay beat me to the response. Hi PhanthomJay!]
 
  • #4
@Phanthom Jay:Everything is downwards, right?
 
  • #5
collinsmark said:
I assume that "F(parallel)" is the component of the gravitational force that is parallel to the ramp? That's fine if it is, I just want to make sure that we're both talking about the same thing.

For this particular problem, is "F(parallel)" in the same direction as the frictional force or in the opposite direction? Draw a free body diagram if that helps.

[Edit: PhanthomJay beat me to the response. Hi PhanthomJay!]
yeah, F(parallel) is the gravitational force parallel to the incline. It's in the same direction of the frictional force, but I'm not sure if it's right to subtract the force of friction from F(parallel), as the example in the book does, because I got the wrong answer.
 
  • #6
Why would you subtract them if they act in the same direction? And once you find the acceleration, you still need to find the distance it travels up the plane.

Hi, Collinsmark!
 
  • #7
I have to sign off.. :zzz:Collinsmark can take it from here, please, thanks.
 
  • #8
phizics09 said:
yeah, F(parallel) is the gravitational force parallel to the incline. It's in the same direction of the frictional force, but I'm not sure if it's right to subtract the force of friction from F(parallel), as the example in the book does, because I got the wrong answer.
Well, in the example in the book was the F(parallel) in the same direction as the frictional force or in the opposite direction?

Base your formulas (many of which you will be deriving yourself) on your own free body diagrams (FBDs), created for the specific problem at hand. Formulas that apply to one problem don't necessarily apply the same way to other problems. That's why FBDs are so important.

On a different note, earlier you asked a question about g being positive or negative. The standard gravity g is always a positive constant. It doesn't mean that acceleration has to be positive though. It depends on how you define positive or negative direction. You get to define this, by the way. For example you could define up as being the positive direction for a problem involving a free-falling object. In that case, since the object is accelerating down, a = -g. But note that g is still positive even though the acceleration isn't. Sometimes you'll need to stick a negative sign in front of g to make the acceleration negative, but that doesn't make g itself is negative.
 
  • #9
PhanthomJay said:
Why would you subtract them if they act in the same direction? And once you find the acceleration, you still need to find the distance it travels up the plane.

Hi, Collinsmark!

I don't know, that's what the example in the books does.
 
  • #10
collinsmark said:
Well, in the example in the book was the F(parallel) in the same direction as the frictional force or in the opposite direction?

Base your formulas (many of which you will be deriving yourself) on your own free body diagrams (FBDs), created for the specific problem at hand. Formulas that apply to one problem don't necessarily apply the same way to other problems. That's why FBDs are so important.

On a different note, earlier you asked a question about g being positive or negative. The standard gravity g is always a positive constant. It doesn't mean that acceleration has to be positive though. It depends on how you define positive or negative direction. You get to define this, by the way. For example you could define up as being the positive direction for a problem involving a free-falling object. In that case, since the object is accelerating down, a = -g. But note that g is still positive even though the acceleration isn't. Sometimes you'll need to stick a negative sign in front of g to make the acceleration negative, but that doesn't make g itself is negative.

Yeah, Fparallel and Ff is in the same direction. So i shouldn't subtract them but add them together?
 
  • #11
phizics09 said:
I don't know, that's what the example in the books does.
Let me ask in a different way. In the example in the book, was an object sliding down a ramp or sliding up the ramp? It makes a difference! The [dry, kinetic] frictional force is always in opposite direction of the velocity.
phizics09 said:
Yeah, Fparallel and Ff is in the same direction. So i shouldn't subtract them but add them together?
There you go. :approve:

But if you don't understand why that is, it might help to review vectors. Remember, forces are vectors and they must be added together like vectors. Think of vectors like arrows, both a magnitude and direction. When adding vectors together, you take the head of the first arrow and put right on the tail of the second. The result is from the tail of the first to the head of the second.
 
  • #12
collinsmark said:
Let me ask in a different way. In the example in the book, was an object sliding down a ramp or sliding up the ramp? It makes a difference! The [dry, kinetic] frictional force is always in opposite direction of the velocity.

The object is going up.:)
 
  • #13
phizics09 said:
The object is going up.:)
Allow me to reword something I said earlier. In a previous post I said, "The [dry, kinetic] frictional force is always in opposite direction of the velocity." That statement is true when an object is sliding on top of a stationary surface (actually, even if the surface isn't stationary, it's still true as long as the object is moving faster than the surface it happens to be sliding on). But the statement might not apply if there are multiple moving objects involved, all sliding around and all moving. Or it might not apply to a given object if that object is stationary, and some other object is sliding along it. So let me make a more general statement in its place.

The direction of the [dry, kinetic] frictional force on an object is opposite the direction of the object's *relative* motion, with respect to whatever object that it is sliding against. The frictional force is always in a direction such that the force tries to stop/reduce the object from sliding (but not necessarily to stop/reduce the object from moving, such as when both objects are moving). Frictional force tries to reduce the relative sliding motion, is what I'm trying to say.

I haven't seen the example in the book, so I can't comment on what's going on there.
 
  • #14
Okay, I get it :) Thankss
 

FAQ: How Far Will the Skateboarder Travel Up the Second Incline?

1. What is an inclined plane problem?

An inclined plane problem is a type of physics problem that involves a flat surface that is at an angle or incline, rather than being completely horizontal. This type of problem often involves calculating the force required to move an object up or down the incline, as well as the velocity and acceleration of the object.

2. How do you calculate the force on an inclined plane?

The force on an inclined plane is calculated using the formula F = mgsinθ, where F is the force, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline. This formula takes into account both the weight of the object and the force required to overcome the incline.

3. What is the difference between a frictionless and a frictional inclined plane problem?

In a frictionless inclined plane problem, the surface of the incline is assumed to have no friction, meaning that no additional force is required to move the object up or down the incline. In a frictional inclined plane problem, the surface does have friction, so an additional force must be applied to overcome this friction and move the object.

4. How does the angle of the incline affect the force required to move an object?

The steeper the angle of the incline, the greater the force required to move an object up or down the incline. This is because a steeper incline means the component of the weight of the object acting against the incline is greater, making it more difficult to move the object.

5. What are some real-life examples of inclined plane problems?

Some real-life examples of inclined plane problems include pushing a heavy object up a ramp, pulling a sled up a snow-covered hill, and using a car jack to change a tire. These situations all involve an object being moved up an inclined surface, and the force required to do so can be calculated using the principles of inclined plane problems.

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