# Truck brakes and the distance they travel

## Homework Statement

Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of friction in the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 7.90° and the coefficient of friction is 0.380. Find the length of a ramp that will stop a 15222.0 kg truck that enters the ramp at 37.0 m/s.

## Homework Equations

E=0.5mv2
Work= Fd
Ffriction = $$\mu$$mgcos$$\vartheta$$

## The Attempt at a Solution

i equated the work equation to the energy one to form Fd=0.5mv2. for F I substituted this equation :
Ffriction = $$\mu$$mgcos$$\vartheta$$ and solved for distance... and i'm not getting the answer would love any help given thanx

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Shooting Star
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## The Attempt at a Solution

i equated the work equation to the energy one to form Fd=0.5mv2. for F I substituted this equation :
Ffriction = $$\mu$$mgcos$$\vartheta$$ and solved for distance... and i'm not getting the answer would love any help given thanx
The truck also gains some gravitational PE, which has to be taken into account.

so would the equation look like Fd= 0.5mv2+ mgh ?? what would h be equal to ??

LowlyPion
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so would the equation look like Fd= 0.5mv2+ mgh ?? what would h be equal to ??
A slight modification.

m*g*h will be what goes to gravity

μ*m*g*cos7.90°*d is what goes to friction

Sin7.90° = h/d

h = d*sin7.90°

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so then wld the overall equation look like ::0.5mv2+mgdsin28.5 = μmgcos28.5d???

LowlyPion
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so then wld the overall equation look like ::0.5mv2+mgdsin28.5 = μmgcos28.5d???
Not quite.

It's conservation of energy. Put yourself in the instant the truck starts up the ramp.

What is the total energy? I know you know it's 1/2*m*v2

Now consider what happens to that energy by the time it stops.

Some goes to the friction of the lovely gravel. Call that Work of Friction. And some goes to increasing the gravitational potential.

KE = PE_g + W_friction

Shooting Star
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μ*m*g*cos28.5*d is what goes to friction

Sin28.5 = h/d

h = d*sin28.5
Err...what is 28.5? Am I missing something very obvious?

LowlyPion
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Err...what is 28.5? Am I missing something very obvious?
Why yes. Something very obvious.

You are missing that this is the angle in another problem that I had helped with elsewhere and unnecessarily carried into this one through total carelessness, in not just writing θ or double checking to see what the θ of this incline was.

Thanks for the catch.

I'll fix it immediately to minimize further confusion.

k i just wanna make sure my form is right ::

0.5*m*v2 = mgdsin7.90 + μmgcos7.90*d .... does that look right??

LowlyPion
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k i just wanna make sure my form is right ::

0.5*m*v2 = mgdsin7.90 + μmgcos7.90*d .... does that look right??
Yes. That looks good.