Force, Velocity, and Ramp Problem?

[cheerspens]

Homework Statement

A runaway truck with a mass of 10,000 kg reaches an emergency pullout ramp moving at 90 mi/hr. As it plows through the deep sand spread over the pullout, it experiences a frictional force of 5000 Newtons opposing its motion. How high up the ramp will it rise?
Note that 1 mile=1609 meters.

Homework Equations

The Attempt at a Solution

So far I've only converted 90 mi/hr to 40.23 m/sec.
I'm totally confused on how to go about solving this problem. Maybe the fact that the ramp at 10 degrees is throwing me off?

 

[pgardn]

Homework Statement

A runaway truck with a mass of 10,000 kg reaches an emergency pullout ramp moving at 90 mi/hr. As it plows through the deep sand spread over the pullout, it experiences a frictional force of 5000 Newtons opposing its motion. How high up the ramp will it rise?
Note that 1 mile=1609 meters.

Homework Equations

The Attempt at a Solution

So far I've only converted 90 mi/hr to 40.23 m/sec.
I'm totally confused on how to go about solving this problem. Maybe the fact that the ramp at 10 degrees is throwing me off?

Have you done conservation of energy in your physics class yet? Or just kinematics and Forces?

 

[cheerspens]

Yes how would you use Energy Conservation to solve this?

 

[pgardn]

Yes how would you use Energy Conservation to solve this?

Do you know the equations for kinetic and potential energy?

 

[cheerspens]

KE would be in the beginning. KE and IE would be in the end.
Now that I know this though, how do you solve for delta X? THat is what you would have to solve for in this case right?

 

[pgardn]

KE would be in the beginning. KE and IE would be in the end.
Now that I know this though, how do you solve for delta X? THat is what you would have to solve for in this case right?

Yes to the all KE in the beginning, but as the truck goes up the ramp it loses KE and converts it to U or PE. If it says the highest the truck rises, this implies the truck comes to a halt momentarily or not. If it was not at its highest it would be moving back down or still moving up the ramp. Now this all assumes the ramp is frictionless. But you are also going to lose some energy to friction.

So do you know the equations for gravitational potential energy and Kinetic energy? And do you know the work friction is performing on the truck?

 

[Maybe_Memorie]

Are we assuming here that before the truck meets the ramp the ground is smooth, and there are no forces acting on him appart from weight and reaction?

I tried solving it with these assumptions, and I got a strange answer.. what's the answer supposed to be?

 

[cheerspens]

Yes to the all KE in the beginning, but as the truck goes up the ramp it loses KE and converts it to U or PE. If it says the highest the truck rises, this implies the truck comes to a halt momentarily or not. If it was not at its highest it would be moving back down or still moving up the ramp. Now this all assumes the ramp is frictionless. But you are also going to lose some energy to friction.

So do you know the equations for gravitational potential energy and Kinetic energy? And do you know the work friction is performing on the truck?

Yes I know GPE and KE.
So as we said before there is KE in the beginning and KE and IE in the end. Would there also be GPE in the end and that gives you how high up the ramp the truck will rise like the question asks? And what do you do with the delta X that is unknown?

Could the answer possibly be 82.57 meters?

 

[pgardn]

Yes I know GPE and KE.
So as we said before there is KE in the beginning and KE and IE in the end. Would there also be GPE in the end and that gives you how high up the ramp the truck will rise like the question asks? And what do you do with the delta X that is unknown?

Could the answer possibly be 82.57 meters?

If you assume that the beginning of the ramp is your zero GPE level, then its all KE initially. This would be the easiest way. Why do you want to assume the truck is still moving at its highest point on the ramp (ie has KE at the end)? At its highest point up the ramp assume it is not moving.
So using the conservation of energy


K = Ug + work done by friction

Does this make sense?

 

[cheerspens]

So the bottom of the ramp would have KE (.5mv²) and the top of the ramp would have GPE (mgh). In order to take friction into consideration would you have Internal Energy (IE) at the top of the ramp as well (F_F$$\Delta$$Xcos$$\theta$$)?
How do you take into account the frictional force of 5000 Newtons because if you use F_F$$\Delta$$Xcos$$\theta$$ what would $$\Delta$$X be?

 

[pgardn]

So the bottom of the ramp would have KE (.5mv²) and the top of the ramp would have GPE (mgh). In order to take friction into consideration would you have Internal Energy (IE) at the top of the ramp as well (F_F$$\Delta$$Xcos$$\theta$$)?
How do you take into account the frictional force of 5000 Newtons because if you use F_F$$\Delta$$Xcos$$\theta$$ what would $$\Delta$$X be?

Just .5mv^2 at the bottom... and mgh and the work done by friction and given off as heat This picture may help you relate X, distance up the ramp at an angle theta and h... I think this route is easiest the way you are thinking.

In other words, all the Kinetic energy of the truck is converted to gravitational potential energy and heat.

 

[cheerspens]

How would I even find x (the length traveled up the ramp) or the height, h, as shown in your picture when all that was give is the angle of the ramp to be 10 degrees?

Would you have an equation that looked like this:
.5mv²=mgh+F_F$$\Delta$$Xcos$$\theta$$

I know that the truck is in motion before the ramp so that is the kinetic energy .5mv². Then as that truck travels up the ramp there is gravitational potential energy mgh because the truck is off the ground. Now I'm not sure if I'm right but I'm thinking my confusion sets in with the 5000 Newtons friction and how to take it into account.
If this was a frictionless ramp you would just set .5mv²=mgh and solve for h to find the height right? So how do I add in the friction?

(Thanks for providing so much help...I really appreciate it. Sorry I'm not getting it yet.)

 

[pgardn]

How would I even find x (the length traveled up the ramp) or the height, h, as shown in your picture when all that was give is the angle of the ramp to be 10 degrees?

Would you have an equation that looked like this:
.5mv²=mgh+F_F$$\Delta$$Xcos$$\theta$$

I know that the truck is in motion before the ramp so that is the kinetic energy .5mv². Then as that truck travels up the ramp there is gravitational potential energy mgh because the truck is off the ground. Now I'm not sure if I'm right but I'm thinking my confusion sets in with the 5000 Newtons friction and how to take it into account.
If this was a frictionless ramp you would just set .5mv²=mgh and solve for h to find the height right? So how do I add in the friction?

(Thanks for providing so much help...I really appreciate it. Sorry I'm not getting it yet.)

So right now we have m, g, h, v, theta, x and the force due to friction. Out of all of these you know m, g, v, theta, force due to friction. The only things missing are the distances, h and x. But h and x are related. I think you might be able to put h in terms of x. Then there will only be one variable, x. This is what you are looking for I believe. I am calling x the distance the truck travels up the ramp based on my picture.

In my picture I am using the force of friction and X as parallel to each other so no need for cosine.

0.5mv^2 = mgh + F*x

So put the h in terms of x and solve for x.

 

[cheerspens]

I got an answer of 54.96 meters. Is that correct?
My work is as follows:

.5mv²=mgh+F_F$$\Delta$$Xcos$$\theta$$
.5(10000)(40.23)²=10000(9.8)h + 5000$$\Delta$$Xcos(10)
8092264.5 = 147240.04h
h=54.96 m

 

[pgardn]

I got an answer of 54.96 meters. Is that correct?
My work is as follows:

.5mv²=mgh+F_F$$\Delta$$Xcos$$\theta$$
.5(10000)(40.23)²=10000(9.8)h + 5000$$\Delta$$Xcos(10)
8092264.5 = 147240.04h
h=54.96 m

Using your numbers... I did not go through changing mph to m/s etc..

0.5(1000)(40.23^2) = 1000(9.8)(sin 10)*x + 5000*x

I put h in terms of x... sin 10 = h/x by my picture...