Conservation of energy of a truck problem

In summary, the conversation is discussing a problem involving a 25 ton truck with no brakes that reaches the bottom of a hill at a speed of 60 mph. The question is how far the truck will travel along a runaway truck ramp that is inclined at a 15 degree angle to the horizontal, assuming no energy losses. The conversation includes a formula for energy conservation and the use of equations for kinetic and potential energy to solve for the distance. However, there is confusion about the conversion of units and the final answer is incorrect.
  • #1
tre2k3
18
0
I tried many ways to perfomr this problem, but i just can't figure out whether or not I am using the right values.

"A 25 ton truck loses it brakes and reaches the bottom of a hill with a speed of 60 mph.
Fortunately, there is a runaway truck ramp which is inclined at angle of 15 degrees to the horizontal."
Assuming no losses, what distance does the truck travel along the runaway truck ramp? (miles) ?

The KE of the truckjust before it goes up the ramp is 6010000 ft-lb.


What do I need to do to get the distance the truck travel along the ramp? The length of the ramp is 1 mile if that means anything.

Heres a formula I've used : KE1+PE1+Work=KE2+PE2+Eloss

Im not sure what force i need to use. I would guess the 6010000 I found.
 
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  • #2
Draw a picture of the hill.
You know how high it will go by using conservation of energy.
At the bottom it will only have kinetic energy (which value you know).
At the highest point it will only have grav. potential energy (which must equal the total energy).
 
  • #4
you can delete this post
 
  • #5
First off, I don't get 6010000 ft*lbs for the Kinetic Energy of the truck, Kinetic energy is measured witht the equation (1/2 m * v^2) where, in the english system, m must be in slugs (lbs/g), and v in (ft/s).

You should look at the type of energy conversion you are doing, you start with a force which is entirely kinetic energy, and you end with no kinetic energy, because your velocity is 0.

Because of the Laws of Conservation of Energy, you cannot simply lose energy, so your energy must have been converted to another form. Because their is no friction acting, the energy must all be transferred to potential energy.

The equation for potential energy is (mgh) where m equals the mass of the truck, g is the acceleration due to gravity (32.2 ft/s) and h is the height above the starting height (the bottom of the hill)

Because these two energies must be equal you can set 1/2mv^2 = mgh and solve for h, then use trigonometry to solve for the distance along the ramp the truck must travel to get to the necessary height.

Hope this helps,

~Lyuokdea
 
  • #6
Lyuokdea said:
First off, I don't get 6010000 ft*lbs for the Kinetic Energy of the truck, Kinetic energy is measured witht the equation (1/2 m * v^2) where, in the english system, m must be in slugs (lbs/g), and v in (ft/s).

You should look at the type of energy conversion you are doing, you start with a force which is entirely kinetic energy, and you end with no kinetic energy, because your velocity is 0.

Because of the Laws of Conservation of Energy, you cannot simply lose energy, so your energy must have been converted to another form. Because their is no friction acting, the energy must all be transferred to potential energy.

The equation for potential energy is (mgh) where m equals the mass of the truck, g is the acceleration due to gravity (32.2 ft/s) and h is the height above the starting height (the bottom of the hill)

Because these two energies must be equal you can set 1/2mv^2 = mgh and solve for h, then use trigonometry to solve for the distance along the ramp the truck must travel to get to the necessary height.

Hope this helps,

~Lyuokdea
6010000 ft*lbs KE i got from dividing 8200000J by 1.36J. They just gave me some conversion factors to work with. I am thinking i should use F*Scos(theta) which equals work.
 
  • #7
I converted my initial KE from ft-lbs to J and had that = to my inital KE. So i converted 6010000 ft-lb to J by multiplying by 1.36J/ft-lb which = to 8200000

I use the equation KE1+PE1+Work=KE2+PE2+Eloss
Since no energy was loss I canceled that out. I canceled out PE1 because the height was 0. So I ended up with KE1+Work=KE2+PE2. Since the ramp is 1 mile long, I converted that to meters by multiplying 1mile x 1609.344m/mile which is just 1609.344.

The values are now
8200000+Fcos(15)s=8200000+1/2(23000)(9.81)(1609.344m)
I came out with s=16000 m
I converted that to miles by dividing by 1609.344 m to get around 9.94 miles for my answer but after doing all that, I go the wrong answer.
 

Related to Conservation of energy of a truck problem

What is the conservation of energy of a truck problem?

The conservation of energy of a truck problem is a physics problem that involves the analysis of the energy of a truck in motion. It is based on the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted.

How is the conservation of energy applied to a truck problem?

In a truck problem, the conservation of energy is applied by considering all the different forms of energy involved, such as kinetic energy, potential energy, and thermal energy. The total energy of the truck at any given moment should remain constant, even as it undergoes changes in speed, elevation, and friction.

What are the factors that affect the conservation of energy in a truck problem?

The factors that affect the conservation of energy in a truck problem include the mass and velocity of the truck, the height and angle of incline of the road, and the type and condition of the truck's brakes and tires. These factors can influence the amount of energy that is transferred or converted throughout the problem.

How is the conservation of energy verified in a truck problem?

In order to verify the conservation of energy in a truck problem, the total energy at the beginning and end of the problem must be calculated and compared. If the total energy remains the same, then the conservation of energy is confirmed to be true. Any discrepancies in the calculations may indicate a violation of the principle of conservation of energy.

What are the real-world applications of the conservation of energy in a truck problem?

The conservation of energy in a truck problem has real-world applications in transportation and logistics, as well as in the design and efficiency of vehicles. It is also applicable in the analysis of accidents involving trucks, as it can help determine the energy involved in the collision and the potential consequences. Additionally, understanding the conservation of energy can aid in the development of more sustainable and energy-efficient transportation systems.

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