Physics Problem

[scalpel179]

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I am having a little trouble with this physics problem. Perhaps someone can enlighten me with their knowledge in this area.

Here is the problem:

A 10 kg block is traveling with a velocity of 10 m/s across a smooth surface when it encounters a rough incline of 30 degrees that rises to a height of 5 meters. If the coefficient of kinetic friction between the block and the inline is 0.3, and the coefficient of static friction is 0.4, what is the final position of the block?

My solution is that it travels up the ramp 7.31 meters and then travels back down the ramp with an acceleration of 1.96m/s^2. this would mean that the block leaves the ramp heading back accross the smooth surface with a velocity of 5.35 m/s.

I don't think that is what the problem is implying, please help.

sincerely,

Scalpel179

 

[notJERRYFALWELL]

Huh? Why would it go back down?

 

[Mr Reddly]

tan(30) > 0.4 = slides back down... no?

In order for the box to remain stationary, this force must be less than the maximum frictional force:

mg sin(q) < Us*mg*cos(q)
or
tan(q) < Us.
where Us = 0.4 and q = 30


edit (for the edit): haha! We're just too fast fer ya! 😛

 

[samurai_lincoln]

Huh? Why would it go back down?

Um, there is this thing called gravity. Perhaps you have heard of it?

On a more serious note, I am bored to death at work and will see what I can do with this.

Edit: dang, someone beat me to it. oh well.

 

[Mr Reddly]

I think the prof messed up. He probably wanted a ligher box, a smaller angle, or a rougher serface on that plane. (edit: forget the comment about a ligher box... that won't matter...oops)

He didn't ask for velocity or anything. He just wanted to know the possition.
If Fs can't stop the box, then it either went up and over, or it went back down the way it came. So to me, the only question is "did the box go higher than 5 meters up or not?"

 

[fatboy_jim]

I too am insanely bored at work and thought I'd give it a shot.

My answer: it gets 6.71m up the hill, and then proceeds to slide back down (didn't bother finding out how fast it returned down the hill). This is a smidge different to your answer, but I think it's safe to say the question is worded horribly. Due to my engineering background, I'll round 6.71m to 7m and I'll round your 7.3m to 7m, and call it good.

Was this an MCAT problem, or what?

 

[stoleyerscrubz]

when i get home from going out tonight I will give it a whack.
a couple of things though:

once the block stops moving due to Uk it then must have a force acting upon it that is greater than Us for it to slide back down. Do not assume it will slide back down just because it is on an incline. Also on first look it seems that the "5m" is extraneous info and you need to decide if Us or Uk should be used.

 

[chiripero]

I too am insanely bored at work and thought I'd give it a shot.

My answer: it gets 6.71m up the hill, and then proceeds to slide back down (didn't bother finding out how fast it returned down the hill). This is a smidge different to your answer, but I think it's safe to say the question is worded horribly. Due to my engineering background, I'll round 6.71m to 7m and I'll round your 7.3m to 7m, and call it good.

Was this an MCAT problem, or what?

How are you getting this answer?

 

[chiripero]

I am having a little trouble with this physics problem. Perhaps someone can enlighten me with their knowledge in this area.

Here is the problem:

A 10 kg block is traveling with a velocity of 10 m/s across a smooth surface when it encounters a rough incline of 30 degrees that rises to a height of 5 meters. If the coefficient of kinetic friction between the block and the inline is 0.3, and the coefficient of static friction is 0.4, what is the final position of the block?

My solution is that it travels up the ramp 7.31 meters and then travels back down the ramp with an acceleration of 1.96m/s^2. this would mean that the block leaves the ramp heading back accross the smooth surface with a velocity of 5.35 m/s.

I don't think that is what the problem is implying, please help.

sincerely,

Scalpel179

I think the block flies off of the ramp. The energy given off by heat because of the friction with the ramp is around 250Joules (Normal Force(mgcos(30)) * Coefficient of kinetic friction * Distance (which is 10 using trig), the energy the block enters the ramp with is 500Joules(1/2mv^2). Thus, the Block flies off.

 

[Premedtomed]

I am having a little trouble with this physics problem. Perhaps someone can enlighten me with their knowledge in this area.

Here is the problem:

A 10 kg block is traveling with a velocity of 10 m/s across a smooth surface when it encounters a rough incline of 30 degrees that rises to a height of 5 meters. If the coefficient of kinetic friction between the block and the inline is 0.3, and the coefficient of static friction is 0.4, what is the final position of the block?

My solution is that it travels up the ramp 7.31 meters and then travels back down the ramp with an acceleration of 1.96m/s^2. this would mean that the block leaves the ramp heading back accross the smooth surface with a velocity of 5.35 m/s.

I don't think that is what the problem is implying, please help.

sincerely,

Scalpel179

initial velocity = 10m/s. Final 0m/s.
Now total force to over come = Fnormal * coeff. of static friction + kinetic friction force.

Now I am lost.

 

[musiclink213]

yuck. physics. i just had an exam on it today on basically that same exact thing. only i'm afraid i can't help you, as physics makes absolutely no sense to me whatsoever, and it seems very illogical. If i want to know if it'll slide back down, then i'll take a block, push it up an incline, let go, and see what happens. A little work never hurt anybody.

 

[LUBDUBB]

Well, fatboy Jim is right, it goes up 6.71 m, and ....However, it really doesn't matter, when it comes back down it will continue to slide indefinately (newton's first) on the smooth surface no matter what its velocity is. If there were friction on the smooth surface, then it would be possible to calculate the final position.

Chiripero, use conservation of energy to solve for 6.71m.

 

[scalpel179]

Thanks for the input. I still calculate the distance that the block travels as 7.31 m. Here is the math. (I agree with the statement that the block will slide back down and then essentially travel forever on the frictionless surface.)

first a FBD.

My formula to solve for acceleration: F-kinetic friction-mg sin 30= ma

solve for a=(f-friction mg- mg sin 30)/m

(10-(.3*10*9.8)-(10*9.8*sin 30))/10

a= -6.84 m/s^2

This value is then substituted into the kinetics equation :

v^2=vo^2+2a(x-xo)

0-10^2/2(-6.84)=7.31m

the problem is then reversed and the block is sliding down the incline plane of 30 degrees. velocity original is zero, acceleration is solved for and equals 1.96 m/s^2. This can be used to solve for a final velocity of 5.35m/s.

Sorry this is difficult to convey over the computer, but it is driving me nuts and I would like to be done with this homework problem.

PS we have not learned conservation of energy yet, nor can I see how joules and energy would have to do with this problem

 

[Mr Reddly]

I'm afraid I can't help much.... been tooooooo long.

However,

My formula to solve for acceleration: F-kinetic friction-mg sin 30= ma

I can see that friction is a force and gravity is a force, but what is F=10? Where did that force come from? That's your 'ma' right?
so
F-kinetic friction-mg sin 30=0
ma-kinetic friction-mg sin 30=0
kinetic friction-mg sin 30= ma

v^2=vo^2+2a(x-xo)

0-10^2/2(-6.84)=7.31m

Are you saying Vo = 10?
I think Vo(x) = 10, but Vo(parallel to ramp != 10.... right?
Wouldn't Vo = 10*cos(30) ?

 

[Mr Reddly]

Chiripero, use conservation of energy to solve for 6.71m.

It's been toooooo long, but I'm curious now... is energy conserved on this? Friction on the plane will give off heat... I think... Or are you taking that into consideration?

 

[LUBDUBB]

Conservation of energy

Work = dKE + dPE d=delta

rearranged as


Initial KE + Initial PE + work = Final KE + Final PE

KE = 1/2mv^2 v-velocity
PE = mgh h =height
work = force x distance = -Friction * D = mgCos(30)(.3)(D)

therefore

1/2m(10)^2 + mg(0) - mgCos(30)(.3)(D) = 1/2m(0)^2 + mg(sin30)(D)



the initial velocity is 10 so we use this for the initial KE.

The box starts at 0 m, so the intial PE=0.

The object reaches its peak on the inclined plane at Velocity =0
therefore the final KE = 0.

D is the distance the box travles up the plane (hypotenuse). Using some
basic trig, the final height of the box is Dsin30.


Sorry if this got too long or confusing...

 

[ih8biochem]

Let me take a whack at this:

the box has an initial kenetic energy of 500 J (1/2 * (10) * (10)^2)
if the box comes to a rest, 150 J of that is lost to friction (500 * 0.3)
Therefore 350 J is converted to potential energy (when the box comes to a rest).
So the maximum height would be 3.5 m (350 J / (10kg) (10m/s^2))
I'm not sure of you can do this but I'm guessing that box travels 3.5/5 up the inclined plane (5 is the height of the ramp)
3.5/5 = 35/50 = 7/10
the ramp (sin 30 = opp/hyp) is 10 meters.
7/10 * 10 meters = 7 meters
Therefore, the ramp travels 7 meters up the ramp.
Wow, see how all the numbers are so easy to calculate. This has gotta be an MCAT problem. 😀

Anyone have any flaws to point out in this calculation?

 

[ih8biochem]

Thanks for the input. I still calculate the distance that the block travels as 7.31 m. Here is the math. (I agree with the statement that the block will slide back down and then essentially travel forever on the frictionless surface.)

first a FBD.

My formula to solve for acceleration: F-kinetic friction-mg sin 30= ma

solve for a=(f-friction mg- mg sin 30)/m

(10-(.3*10*9.8)-(10*9.8*sin 30))/10

a= -6.84 m/s^2

This value is then substituted into the kinetics equation :

v^2=vo^2+2a(x-xo)

0-10^2/2(-6.84)=7.31m

the problem is then reversed and the block is sliding down the incline plane of 30 degrees. velocity original is zero, acceleration is solved for and equals 1.96 m/s^2. This can be used to solve for a final velocity of 5.35m/s.

Sorry this is difficult to convey over the computer, but it is driving me nuts and I would like to be done with this homework problem.

PS we have not learned conservation of energy yet, nor can I see how joules and energy would have to do with this problem

Ahh.. this is a homework problem. I should've looked at this earlier. I think there's two forces acting to slow down the box, friction and gravity. You just have friction, that's probably why your calculation came out slightly higher than mine. Try doing it again kinematically with gravity factored in. But wait... I think gravity is already factored in in your value for a. Now I'm confused.

 

[ih8biochem]

ok, I'm determined to get this...
you use [mg cos 30 * kinetic friction co] to get force 1.
For gravity, you use [mg sin 30]

total deceleration of about -7.45 m/s^2

plug and chug and I get 6.71 meters.

something still doesn't seem right though...

 

[elias514]

I think I have the solution you're looking for, but the semantics of the question remain somewhat of a problem. Here are the steps of the solution:

1. Use Newton's second law (F=ma). Specifically, F=ma=(force on box due to gravity)+ (kinetic frictional force). This formula can be rewritten, substituting numerical values as:

(10 kg)(acceleration)=(10 kg)(-9.8 m/s^2)(sin 30) + (10 kg)(-9.8 m/s^2)(sin 30)(.3)

Solving for acceleration, a=-6.37 m/s^2. Thus, every second, the box will lose 6.37 m/s of velocity on the ramp.

2. Use a kinematics equation to find the displacement along the ramp. The equation is (final velocity)^2=(initial velocity)^2 + 2(acceleration)(displacement). Substituting numbers into the equation and solving for displacement, one gets the following:

(0-100)/2(-6.37)=7.85 meters

Thus, the box travels 7.85 meters up the incline before coming to rest. This could be the answer to the question, assuming that the question writer wanted the final position on the ramp and did not care about what happens next (i.e., the box begins to move back down the ramp).

3. Find the potential energy of the box. Using trigonometric relationships, one can find the height that corresponds to a displacement of 7.85 meters; this height is 3.925 meters. So the initial potential energy of the box on the inclined plane is given by the formula for gravitational potential energy--PE=mgh.

(10 kg)(9.8)(3.925)=385.65 Joules

So the box initially has approximately 385.65 Joules, which would all be converted into kinetic energy, were it not for the presence of friction, which will cause some of this initial potential energy to be "lost" as heat.

4. Find out how much work the kinetic frictional force does on the box. Work is equal to the force times the distance.

work=(kinetic frictional force)(7.85 m)=(10)(9.8)(sin 30)(.3)(7.85)=115.395 Joules

So the kinetic frictional force will sap 115.395 Joules of potential energy from the box. This means that the box will exit the ramp with 270.255 Joules of kinetic energy. Using the kinetic energy formula, one can easily find the velocity that corresponds to this energy value: namely, 7.35 m/s.

Since there is no friction on the smooth surface. The box will continue to move indefinitely.

Hope this helps!

 

[LUBDUBB]

Yeah you guys are right...

Scalepl,

You can try drawing free standing diagrams to see all the forces acting on the box. When the box is going up the plane, two forces are opposing the motion, Friction and gravity. So

Force friction + Force grav = ma

mgCos(30)(.3) + mgSin(30) = ma

a=7.45

Put into your kinematics eq. v^2 = v0^2 + 2a (dx)

and you'll get 6.71m

 

[HessExpress]

lubdubb's got it, it's all in the free-body diagrams - make one for every problem you do in the force chapter.

 

[scalpel179]

Thanks for all of those who helped. We just learned consernation of energy, along with work, potential energy, and kinetic energy. This certainly would have been an easier method to solve this problem. Nontheless, thanks for all your valant efforts. I did finish my homework and LUBBDUBB had it correct. travels up 6.71m, overcomes static friction than travels back down the incline continuing indefinitely.

Thanks again