Inclined Planes, finding time from acceleration at an angle

[ericcy]

Homework Statement

Starting from rest, a cyclist coasts down the starting ramp at a professional biking track. If the ramp has the minimum legal dimensions (1.5 m high and 12m long), find

b) the acceleration of the cyclist if all sources of friction yield an effective coefficient of fricition of 0.11(solved this, answer is 0.1554m/s^2, this provides context for next question)

c) the time taken to reach the bottom of the ramp, if friction acts as in (b)

Relevant Equations

Fgx=Sin(theta)mg
Fgy-Cos(theta)mg
ma=Fnet
Ei=Ef

Because the friction is the same in both parts, the calculated acceleration from (b) should be the same for (c)

I knew I could find Vf, and thought I could do it with an energy equation

Ei=Ef
mgh=1/2mv^2
gh=1/2v^2
(2)9.81(1.5)=1/2v^2(2)
(square root)29.43=(square root)v^2
v= 5.424

Then kinematics to determine the time
Vf=Vi+at
5.424/1.554=0.1554t/1.554
t=34.9s

Pretty sure that this should work, but maybe it won't for inclined planes? I'm not sure. It could be that I didn't include the actual energy lost from friction (ff=1.0706)? I was thinking that I could also do it by using the hypotenuse of the triangle formed (12M) as distance for another kinematics equation, but why won't the other method work that I showed above?

The textbook answer is 12s

Sorry for the bunch of questions recently, I have a unit test tomorrow and just want to make sure I understand everything. Thanks guys.

 

[gneill]

I knew I could find Vf, and thought I could do it with an energy equation

Ei=Ef
mgh=1/2mv^2
gh=1/2v^2
(2)9.81(1.5)=1/2v^2(2)
(square root)29.43=(square root)v^2
v= 5.424

Problem is, friction is "steeling" energy from the system as the bike moves. So gravitational PE won't entirely be converted to KE. Some will be lost as heat due to the friction.

Can you think of another approach where the effective coefficient of friction is taken into account?

 

[ericcy]

Problem is, friction is "steeling" energy from the system as the bike moves. So gravitational PE won't entirely be converted to KE. Some will be lost as heat due to the friction.

Can you think of another approach where the effective coefficient of friction is taken into account?

Could I do d=Vit+1/2at^2?

I could use the length from the dimensions, 12m. Would that be the proper way to tackle this question?

 

[gneill]

Could I do d=Vit+1/2at^2?

I could use the length from the dimensions, 12m. Would that be the proper way to tackle this question?

You're not given the acceleration outright, so that would not be helpful.

Suppose for a moment that you know the kinetic friction coefficient, the slope angle, and the mass of the object (bike and rider) on that slope. How would you resolve the net down-slope force acting on the object? (note that any truly unknown values can be represented by variables at this point. So take the mass of the "object" to be M. You'll see that it cancels out in the end, so you never actually have to have a value for it...)

 

[ericcy]

You're not given the acceleration outright, so that would not be helpful.

Suppose for a moment that you know the kinetic friction coefficient, the slope angle, and the mass of the object (bike and rider) on that slope. How would you resolve the net down-slope force acting on the object? (note that any truly unknown values can be represented by variables at this point. So take the mass of the "object" to be M. You'll see that it cancels out in the end, so you never actually have to have a value for it...)

The down-slope force acting on the object would be fgx, so mgSin(theta)= 1.226m Fgy is mgCos(theta)= 9.733m

Fn=Fgy so Fn= 9.733m
Ff=Fn(mu)
Ff=9.733m(0.11)= 1.0706
Ff=1.0706m

So the down-slope force would be 1.226m while the upward force is 1.0706m

Not sure how this can help me solve for time? might need more hints lol.

 

[gneill]

Okay, so synthesizing what you've presented so far using symbols only (a very good thing to do, by the way!), after dividing through by m (since a = F/m),

$a = g \sin(\theta) -\mu g cos(\theta)$

Knowing the acceleration and the distance, can you find the time?

 

[ericcy]

Okay, so synthesizing what you've presented so far using symbols only (a very good thing to do, by the way!), after dividing through by m (since a = F/m),

$a = g \sin(\theta) -\mu g cos(\theta)$

Knowing the acceleration and the distance, can you find the time?

All I can think of is that kinematics equation from earlier. We don't have velocity so that's the only way I can think of.

 

[gneill]

All I can think of is that kinematics equation from earlier. We don't have velocity so that's the only way I can think of.

Now you have a value for the acceleration, right? Perhaps another kinematic equation might come to mind involving distance, acceleration, and time?

 

[ericcy]

Now you have a value for the acceleration, right? Perhaps another kinematic equation might come to mind involving distance, acceleration, and time?

I can't think of it. Maybe d=1/2at^2?

 

[gneill]

I can't think of it. Maybe d=1/2at^2?

That looks very promising. Give it a try!