Newton's Second Law Incline (constant accel)

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Homework Help Overview

The discussion revolves around applying Newton's Second Law to a problem involving a block on an incline with constant acceleration. The scenario includes a block with a mass of 5.0 kg, an incline angle of 37 degrees, and a coefficient of kinetic friction of 0.50. Participants are tasked with determining the force required to move the block up the incline at a constant speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss breaking down the forces acting on the block into components along the y-direction and x-direction. They explore the relationship between the net force, frictional force, and gravitational force acting on the block. Questions arise regarding the correct application of Newton's Second Law and the signs of the forces involved.

Discussion Status

The discussion has progressed with participants providing hints and guidance to clarify the relationships between the forces. Some participants have successfully calculated the frictional and gravitational forces, while others are working through the implications of these calculations on the net force required to maintain constant speed. There is a recognition of the need to carefully consider the direction of forces.

Contextual Notes

Participants are navigating the complexities of force directionality and the implications of constant speed on net force. There is an acknowledgment of potential confusion regarding the application of forces in the context of the incline.

Dakren12
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Newton's Second Law...Incline (constant accel)

Homework Statement



Given: mass of block of 5.0 kg
incline angle is 37theta
coeff kinetic friction is 0.50

What is the force, directed up the inline, required to move the block at constant speed up the incline?

The Attempt at a Solution



So i tried to break it up... to the y-dir and x-dir.

the forces for y-dir. I have Fn and (mg*cos37)
the forces for x-dir. I have fk and (mg*sin37)
I have answer saying its 50.. but i can't seem to figure it out!
 
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Dakren12 said:
the forces for y-dir. I have Fn and (mg*cos37)
the forces for x-dir. I have fk and (mg*sin37)
So far, so good. What's fk equal to?

Hint: What net force must act on the block if it's to move at constant speed?
 


so Fnet > fk?

and fk = mu k * Fn...

so... fk = (.5)(mg*cos37)?
...
so confusing.
 


Dakren12 said:
so Fnet > fk?
No. What's does Newton's 2nd law tell you about net force and acceleration?

and fk = mu k * Fn...

so... fk = (.5)(mg*cos37)?
Good!

There are three forces acting parallel to the incline: Friction; gravity (one component of it); and the applied force up the incline (which is what you're trying to find). You can calculate the first two. And, given my hint, you can then solve for the third.
 


sigh... i think I'm over thinking it...

so friction is approx 20
gravity is about 30...
the unknown applied force x

would i still use f=ma?

so something like (20+x)...minus 30 = 0?
...
hopeless ..
 


Dakren12 said:
so friction is approx 20
gravity is about 30...
the unknown applied force x
Good!

would i still use f=ma?
Sure.

so something like (20+x)...minus 30 = 0?
Almost. Careful with signs. What direction does each force act? Hint: Give anything that acts up the incline a positive sign and anything that acts down the incline a negative sign. Rewrite your equation accordingly.
 


wait a min... the fk here is acting downward huh? ... since we're pushing it upwards...

so (20 + 30) - x = 0
x = 50...!

hmm... didnt think to apply fk going downwards... haha.
 


There you go... Not so bad after all, eh?
 


hahah! sweet. i solved the next one as well.. which is pretty much the question in reverse! yay! thanks Doc.
 

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