Highest velocity reached on a hill

[Drizzy]

Homework Statement

Alexander sled down a hill and on a snow-covered horizontal leveled ground. He wanted to calculate the highest velocity that the sled can reach. The hill is 3 meters high and 6 meters long. When Alexander sled down the hill he traveled 12,5 meters on the snow-covered ground. Alexander then makes the further assumption that the frictional force is 85% of frictional force on the ground.

Calculate the highest velocity that can be reached by the sled.

The Attempt at a Solution

I know that the highest velocity is at the bottom of the hill. Basically when he is about to touch the ground. And I am assuming that I have to calculate the acceleration on the hill and then on the ground so that I can solve the problem. I also know that I am going to use this formula: v^2 - v0^2 = 2as

 

[PeroK]

Homework Statement

Alexander sled down a hill and on a snow-covered horizontal leveled ground. He wanted to calculate the highest velocity that the sled can reach. The hill is 3 meters high and 6 meters long. When Alexander sled down the hill he traveled 12,5 meters on the snow-covered ground. Alexander then makes the further assumption that the frictional force is 85% of frictional force on the ground.

Calculate the highest velocity that can be reached by the sled.

The Attempt at a Solution

I know that the highest velocity is at the bottom of the hill. Basically when he is about to touch the ground. And I am assuming that I have to calculate the acceleration on the hill and then on the ground so that I can solve the problem. I also know that I am going to use this formula: v^2 - v0^2 = 2as

It's not clear what you're assuming regarding friction. Is it that the coefficient of friction is 0.85 on the ground?

 

[Drizzy]

no! The frictional force on the hill is 85% of the frictional force that is on the ground. it doesn't say how much the frictional constant is :/

 

[PeroK]

no! The frictional force on the hill is 85% of the frictional force that is on the ground. it doesn't say how much the frictional constant is :/

Okay, so why not assume the frictional forces are ##F## and ##0.85F## and see what you get?

 

[Drizzy]

i don't know how to solve the problem. Where do I start?

 

[PeroK]

i don't know how to solve the problem. Where do I start?

Start at the top of the hill! You know the height, length of hill, ##g##, ##0.85F##. That's enough to express the velocity at the bottom of the hill in terms of these variables.

 

[Drizzy]

okay but I need the angle to calculate the gravitational force

Cuz the sled is tilted downwards

 

[PeroK]

okay but I need the angle to calculate the gravitational force

Cuz the sled is tilted downwards

If you know the height and the length, then you can work out the angle, surely?

 

[Drizzy]

which angle do I need? the top or the bottom one? Sorry for asking so much :(

 

[PeroK]

which angle do I need? the top or the bottom one? Sorry for asking so much :(

You don't need the angle if you think about energy. I suggest that's a better approach in any case.

 

[Drizzy]

okay mgh but I don't have the mass of the sled

 

[PeroK]

okay mgh but I don't have the mass of the sled

So what? Leave it as m.

It stands to reason that if the problem is solvable, the mass is not relevant.

 

[Drizzy]

(g=10)
Energy = 30*m

(mv^2)/2 = 30*m
v^2 = 60
v = sqrt (60)

Is this right?

 

[PeroK]

(g=10)
Energy = 30*m

(mv^2)/2 = 30*m
v^2 = 15
v = sqrt (15)

Is this right?

What about the friction?

 

[Drizzy]

oh I forgot...

the friction on the hill is not equal to the force because the velocity is not constant. the friction = normal force * mu
F(friction)=mu * F(normal)

Now I need the angle

 

[PeroK]

oh I forgot...

the friction on the hill is not equal to the force because the velocity is not constant. the friction = normal force * mu
F(friction)=mu * F(normal)

Now I need the angle

You "know" the frictional force. It's 0.85F. Think energy again: work done by friction.

 

[Drizzy]

but its not 0.85F it is 0,85 * the friction of the ground . which i don't know what to call

 

[PeroK]

but its not 0.85F it is 0,85 * the friction of the ground . which i don't know what to call

This is why you learn algebra in maths! If you always had the numbers, you'd only need a calculator. The magnitude of the force is also not relevant: only the relationship between the friction on the slope and on the ground. We'll call it ##F## on the ground, hence ##0.85F## on the slope.

Now you have two variables: ##m## and ##F##.

You work though the problem using these as variables (isn't algebra great?) and look for the moment when you can cancel them out. Like you did with ##m## when you looked at motion down the slope without friction. You worked out the energy, potential and kinetic, then canceled the ##m## to get the velocity. This is the same strategy but it's a bit harder to get to the point where you can cancel out ##F## and ##m## and solve for ##v##.

This is real physics, by the way, which putting numbers into a calculator is not. Enjoy it if you can!

 

[Drizzy]

I don't know how to use the friction in an equation. We have a force down the hill and the friction and they are opposite of each other. So we have F=F1-Ffriction

F = m * a

F1 is the force that makes the sled go downwards.

m*a=F1-0,85F

Is this right so far?

 

[PeroK]

I don't know how to use the friction in an equation. We have a force down the hill and the friction and they are opposite of each other. So we have F=F1-Ffriction

F = m * a

F1 is the force that makes the sled go downwards.

m*a=F1-0,85F

Is this right so far?

Yes, but think Energy. It's much simpler when you have a force and distance. How much energy is lost to friction?

 

[Drizzy]

force times distance is work.

the energy lost to friction is 6*0,85F times the downward force

 

[PeroK]

force times distance is work.

the energy lost to friction is 6*0,85F times the downward force

Yes. So, now you can try to express ##v^2## in terms of ##m## and ##F## using what you did earlier with PE and KE and what you have now in terms of energy lost.

 

[Drizzy]

v^2 - v0^2 =2as

v^2 = 60

60 - v0^2 =2a * 6

 

[PeroK]

v^2 - v0^2 =2as

v^2 = 60

60 - v0^2 =2a * 6

Maybe this problem is a bit advanced for you? Do you understand how to relate potential, kinetic and energy lost to friction?

 

[Drizzy]

no we haven't learned that yet :/ I can post my teachers answer and maybe you can help me understand it.

http://imgur.com/C8zIz2o

 

[PeroK]

no we haven't learned that yet :/ I can post my teachers answer and maybe you can help me understand it.

http://imgur.com/C8zIz2o

It looked easier using Energy, but can equally do it using forces and acceleration. The first thing you need is the angle (or, at least, the sine of the angle).

 

[Drizzy]

the angle is Cos^-1 3/6 which is 60 degrees.

 

[PeroK]

the angle is Cos^-1 3/6 which is 60 degrees.

You sure about that?

 

[Drizzy]

yeah the top angle is 60 degrees i believe

 

[PeroK]

yeah the top angle is 60 degrees i believe

Why would you describe the slope of an incline using the top angle? That means a slope of 89° would be almost horizontal. In any case, you always describe an incline using the angle at the bottom, which means a 1° slope is almost horizontal and an 89° slope almost vertical.

 

[Drizzy]

if the top is 60 then the bottom is 30 degrees!

 

[Drizzy]

what do I do next?

 

[PeroK]

what do I do next?

You need to start doing more of the work on this! I thought your idea was to calculate the acceleration down the slope?

 

[Drizzy]

okay so the force down is F1 and the "resulting" force is:

F1-force of friction= m*a

F1= sin(30)*mg

I get that part but then my teacher wrote (2) that the frictional force is equal to m*a2

w8... i think I am getting it.. so on the horizontal fround there won't be a force forward so the only force is the frictional force`?

 

[PeroK]

okay so the force down is F1 and the "resulting" force is:

F1-force of friction= m*a

F1= sin(30)*mg

I get that part but then my teacher wrote (2) that the frictional force is equal to m*a2

w8... i think I am getting it.. so on the horizontal fround there won't be a force forward so the only force is the frictional force`?

You seem to have a habit of writing one line and then stopping. You don't finish what you start. The net force gives you acceleration, which will give you ##v^2##. You need to keep going. Not just write one equation then stop.