Vector Diagram for Frictionless Inclined Plane

[AnthonyC]

I am supposed to take this:
http://www.recklesscaution.com/original.jpg

and draw the vector diagram of the forces acting on the object.

So if it is frictionless, I believe this is all I would need:
http://www.recklesscaution.com/vector.jpg

Is that correct?

 

[Nylex]

Yes, although you should probably split the weight into components.

 

[AnthonyC]

Meaning having a mgsin(25) coming from the box down the slope, and a mgcos(25) coming perpendicular from the bottom of the box through the slope? (If that makes sense)

 

[Nylex]

Yep .

 

[AnthonyC]

Thankyou!

Also, to solve for the kinetic energy at the bottom, can I correctly assume that since it is frictionless, the system is closed, so the law of conservation of mechanical energy would indicate that the Potential Energy at the beginning would be equal to the Kinetic Energy at the bottom?

So, if the mass is 2 kg, then:

Ek initial = 0
Ep initial = mgh = 2(9.81)10.56 = 207.1872 J

Ek final = Ep initial
Ek final = 207.1872 J

And for the velocity at X, this is correct, right?

Ek final = Ep initial
½mv2 = mgh
v = √(2gh) = √(2(9.81)10.56) = √207.1872 = 14.394 m/s

 

[learningphysics]

Your answer is correct Anthony.

 

[AnthonyC]

Ok, now I'm stuck at how to find the velocity at position Y. They do not indicate the distance between X and Y.

So I know you cannot tell me how to do it, but can somebody point me in the right direction?

The acceleration should will decrease while on the horizontal plane, but I'm not sure how to do any part of this equation.

 

[learningphysics]

Ok, now I'm stuck at how to find the velocity at position Y. They do not indicate the distance between X and Y.

So I know you cannot tell me how to do it, but can somebody point me in the right direction?

The acceleration should will decrease while on the horizontal plane, but I'm not sure how to do any part of this equation.

Is the horizontal plane frictionless? What are the forces acting on the object when it is at X? What are the forces in the horizontal direction?

 

[AnthonyC]

They do not indicate if the horizontal plane is frictionless, they say:

"An object of mass m=2.00kg is released from the top of a frictionless inclined plane that makes an angle of 25degrees wiht the horizontal"

It hits position X with a velocity of 14.394 m/s, and the acceleration along the slope is 4.146.

So should I simply assume that there is no friction along the horizontal?

Even still, I'm unsure how to figure Y without knowing hte distance between X and Y.

 

[learningphysics]

They do not indicate if the horizontal plane is frictionless, they say:

"An object of mass m=2.00kg is released from the top of a frictionless inclined plane that makes an angle of 25degrees wiht the horizontal"

It hits position X with a velocity of 14.394 m/s, and the acceleration along the slope is 4.146.

So should I simply assume that there is no friction along the horizontal?

Even still, I'm unsure how to figure Y without knowing hte distance between X and Y.

If friction is 0 (from X to Y), then the net force in the horizontal direction is 0. Hence the horizontal acceleration is 0. If acceleration is 0, then the velocity doesn't change (v2=v1 + at. Since a=0, so v2=v1). And at Y, the velocity is still 14.394 m/s.

 

[AnthonyC]

Ok, thankyou, that makes sense to me.

So from the object's initial position at the top of the incline, the net force on the object would simple be:

N = mg cos25 = 2(9.81)cos25 = 17.78 N

Is that correct?

 

[learningphysics]

Ok, thankyou, that makes sense to me.

So from the object's initial position at the top of the incline, the net force on the object would simple be:

N = mg cos25 = 2(9.81)cos25 = 17.78 N

Is that correct?

No. What is the sum of the forces perpendicular to the incline (that gives the component of the net force perpendicular to the incline... it is 0 since the normal force cancels the component of gravity perpendicular to the incline).

What is the sum of the forces parallel to the incline... there is only one force... that is the component of gravity parallel to the incline.

The component of gravity along the incline is the net force acting on the object.

 

[AnthonyC]

I apologize if I don't understand completley, but it should instead be mgsin25 = 2(9.81)sin25 = 8.292 ?

 

[learningphysics]

I apologize if I don't understand completley, but it should instead be mgsin25 = 2(9.81)sin25 = 8.292 ?

Yes, that's the right answer.

No need to apologize.