Projected Block up Frictionless Plane: Solving (a) - (c)

[th77]

A block is projected up a frictionless plane with initial speed v0 = 3.50 m/s. The angle of incline is theta = 32 degrees. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it bets back to the bottom?
I read through the solution to this problem, but there are a few things I don't understand:
(1) Why did they use the positive x direction to be down the incline?
(2) It said the positive x direction is in the direction of the
acceleration. How is acceleration down when the problem is
talking about the block moving up the plane?
(3) It said the x component of Newton's second law is mg sin
theta = ma . I don't understand how they got this.

 

[Doc Al]

(1) Why did they use the positive x direction to be down the incline?

The choice of positive direction is arbitrary, as long as you stick to one choice throughout the problem. Using this choice, what's the sign of the initial velocity?

(2) It said the positive x direction is in the direction of the
acceleration. How is acceleration down when the problem is
talking about the block moving up the plane?

Don't confuse the direction of motion with the direction of the acceleration. After all, the block goes both up and back down the plane. The acceleration, due to the net force acting down the plane, is always down the plane. (Compare this with tossing a ball in the air. It's velocity is up, then zero, then down--yet the acceleration (which is the rate at which the velocity changes) is always the same: 9.8 m/s^2 down.)

(3) It said the x component of Newton's second law is mg sin
theta = ma . I don't understand how they got this.

The weight acts down. mg sin(theta) is the component of the weight in the x-direction. (Draw a picture and use some trig.)

 

[sanitykey]

The weight of the block is the force opposing its motion and is acting vertically downwards, you can take acceleration due to gravity to be either positive or negative for either direction so long as you are consistent with your workings. They probably took the positive x-direction to be down the inclined plane using the logic that going down the plane would be working with gravity.

Acceleration down the plane is another way of saying decceleration up the plane and although the question states the block is fired upwards there is no force acting upwards after the initial firing, therefore the block is slowing down so it is deccelerating.

Force = Mass*Acceleration
Weight = Mass*Gravity
Acceleration due to gravity on Earth is about 9.81ms^-2

Weight is a force so they are replacing F in F=ma with mg=ma however because the block is on an inclined plane the full force isn't at work instead a component of the force is, the force component, or weight component is mg*sin theta.

 

[th77]

The choice of positive direction is arbitrary, as long as you stick to one choice throughout the problem.

So then they could've chosen positive x direction to be up the plane? Is there any advantage to choosing positive x to be down the plane?

Using this choice, what's the sign of the initial velocity?

I'm guessing the initial velocity is negative because it's going up the incline plane, which here is the negative x direction.

Don't confuse the direction of motion with the direction of the acceleration. After all, the block goes both up and back down the plane. The acceleration, due to the net force acting down the plane, is always down the plane. (Compare this with tossing a ball in the air. It's velocity is up, then zero, then down--yet the acceleration (which is the rate at which the velocity changes) is always the same: 9.8 m/s^2 down.)

My concept of acceleration is shady. So when they say the direction of the accleration is down the incline they mean that acceleration is always positive since down the plane is the positive x direction?

 

[Doc Al]

So then they could've chosen positive x direction to be up the plane?

Sure.

Is there any advantage to choosing positive x to be down the plane?

None obvious to me. (It's pretty common to use "to the right" as positive; Does this plane slope down to the right, by any chance?)

I'm guessing the initial velocity is negative because it's going up the incline plane, which here is the negative x direction.

Right.

My concept of acceleration is shady. So when they say the direction of the accleration is down the incline they mean that acceleration is always positive since down the plane is the positive x direction?

Right.

 

[th77]

None obvious to me. (It's pretty common to use "to the right" as positive; Does this plane slope down to the right, by any chance?)

No, down to the left.

Would it be correct to say that acceleration is an abstract concept? Meaning that it has nothing to do with the direction of an object's motion.

 

[Doc Al]

I don't know what you mean by "abstract concept". Acceleration has to do with the rate at which an object's velocity changes. In this case, the acceleration is a constant, but the velocity (obviously) is not.

You might find this tutorial helpful: https://www.physicsforums.com/showthread.php?t=95426

 

[th77]

Thanks very much for you help!