Newton's Second Law Incline (constant accel)

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To determine the force required to move a 5.0 kg block up a 37-degree incline at constant speed, the problem involves analyzing forces in both the y-direction and x-direction. The frictional force (fk) is calculated using the coefficient of kinetic friction and the normal force, leading to fk = 0.5 * (mg * cos(37)). The gravitational force acting down the incline is mg * sin(37), and the net force must equal zero for constant speed, resulting in the equation (fk + gravity) - applied force = 0. The final calculation reveals that the applied force needed is 50 N, confirming the solution.
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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