What Are the Forces Acting on Boxes on Inclined Planes?

[kevinlikesphysics]

3. A 15 kg box slides down a frictionless ramp of 25 degrees. What is its acceleration?
m/s/s

What is the normal force of a 35 kg box sitting on a slope of 14 degrees? (assume the box is not moving).
N

A 45 kg box is sliding at a constant speed down a ramp of 28 degrees.

What is the frictional force on the box?

What is the coefficient of sliding friction?

After an initial push, a box goes up a frictionless incline of 20 degrees.

What is its rate of deceleration?
m/s/s

A 75 kg skier starts at the top of a 15 m incline at an angle of 25 degrees.
Assuming the coefficient of friction is .15, what is his acceleration?).

m/s/s

What is his speed at the bottom of the hill?
m/s

can someone do those for me just so i can figure out the logic like with the equations id ont get it at all thanks

 

[arildno]

The principle used is called Newton's 2.law of motion, which gives rise to an energy balance sheet often more convenient to work with.

As for how to do this:
Start by decomposing your forces in the tangential and normal directions.

 

[kevinlikesphysics]

i don't know how to do it that why I am asking my teacher is the worst teacher ever she's a nerd who is smart as s*** but can teach for her life and I am really frustrated right now I've been told to use Newtons 2nd law like 20 times now and i don't know how to use it i hae the equations i just don't know how t apply it so i can get the answer

 

[arildno]

Okay, suppose your incline has a positive angle \theta[/tex] to the horizontal.<br /> That means that the unit tangent vector \vec{t} up along the incline must be \vec{t}=\cos\theta\vec{i}+\sin\theta\vec{j} where \vec{i},\vec{j} are unit vectors in the horizontal and vertical directions, respectively. Agreed so far?<br /> <br /> Furthermore, the unit normal \vec{n} with positive vertical component must therefore be \vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{j}<br /> Agreed?<br /> <br /> But now, we can, if we wish, write \vec{i}=\cos\theta\vec{t}-\sin\theta\vec{n}, \vec{j}=\sin\theta\vec{t}+\cos\theta\vec{n}<br /> Verify this!<br /> <br /> This is basically what you need to proceed..

 

[arildno]

For example, the force of gravity, given by -mg\vec{j}
may now be rewritten as -mg\vec{j}=-mg(\sin\theta\vec{t}+\cos\theta\vec{n})

 

[kevinlikesphysics]

i got 1 and 2 but the rest i keep getting wrong


i don't know hwo to get Fk abd Uk

 

[arildno]

Post your ideas for one of the problems, say number 3.

 

[kevinlikesphysics]

i don't know i have

A 45 kg ,a ramp of 28 degrees.

What is the frictional force on the box? ...

coeficient of friction?...

i tried to use ma =mgsin0 - ukFn = mgsin0-ukmgcos0

and got 389.37 for Fk

and i need fk to get coeeficient so i stoped