A truck with mass m has a brake failure while going down an icy mountain road of

Question

A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle (Figure 1) . Initially the truck is moving downhill at speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle . The truck ramp has a soft sand surface for which the coefficient of rolling friction is r.

Part A

What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.

Express your answer in terms of m, , v0, L, g, and r.

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Figure 1 of 1

A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle (Figure 1) . Initially the truck is moving downhill at speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle . The truck ramp has a soft sand surface for which the coefficient of rolling friction is r.

Part A

What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.

Express your answer in terms of m, , v0, L, g, and r.

SubmitMy AnswersGive Up

Incorrect; Try Again; 3 attempts remaining

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Figure 1 of 1

Explanation / Answer

three points at which we can formulate equations for energy:

P is the point where the careening truck's velocity is V0;
Z is the zero-point, the lowest elevation, where the truck starts up the ramp;
S is the point where the truck stops.

Let's look at the kinetic energy KE, the potential energy PE, and the work energy WE required to get to each point; if I'm thinking correctly, we can apply the Conservation of Energy law at all 3 points. That approach has a chance of being able to calculate x, because x is a factor of the energy at point S.


Equations and Steps:

First, formulate an equation for the total mechanical energy at point P:

(2) Eptot = KEp + PEp (this is our initial reference point, so neglect WE required to get there)

= (½ * m * V0²) + (m * g * (L * sin a))

Next, the total energy at Z:

(3) Eztot = KEz + PEz + WEz

PEz = 0, because that's our elevation reference point for PE
WEz = 0, because there is no work done against friction in getting from P to Z (it's a frictionless path)

Therefore,

(4) Eztot = KEz

= (½ * m * V²),

where V is the velocity of the truck at point Z.


Last, the mechanical energy at S, the stopping point of the truck:

(5) Estot = KEs + PEs + WEs

KEs = 0 because the truck is stopped;

(6) PEs = m * g * (x * sin b)

Now, what is WEs, the work expended in getting from Z to S? It is work done against the rolling friction force Frf. I looked at reference 1 and found this formula for Frf:

(7) Frf = Crr * W,

where:
W = the normal force, which is the normal load on the axle (system) = m * g;
Crr = the dimensionless Coefficient of rolling friction;
Frf = the force on the axle in the direction of travel and perpendicular to W

In our case, since the truck's motion is at an angle b to the truck's weight, the weight normal to the ramp is W * cos b, so:

(8) Frf = Crr * (m * g * cos b)

and the work done against that force (F X distance) is:

(9) WEs = Frf * x

= (Crr * (m * g * cos b)) * x

Substituting (6) and (9) in (5):

(10) Estot = KEs + PEs + WEs

= 0 + (m * g * x * sin b ) + ((Crr * m * g * cos b) * x )


Now to recap: We have expressions for Etot at all 3 points:

(11) Eptot = (½ * m * V0²) + (m * g * (L * sin a))

(12) Eztot = (½ * m * V²)

(13) Estot = (m * g * x * sin b ) + ((Crr * m * g * cos b) * x )

Inspecting these equations, we see that the only "stray" variable is the V in (12), the velocity at Z. Here's where Conservation of Energy rides in to save us: If I've analyzed the situation correctly, then

(14) Eptot = Eztot = Estot

So let's just ignore middle Eztot and say that Estot = Eptot, or:

(15) (m * g * x * sin b ) + ((Crr * m * g * cos b) * x ) = (½ * m * V0²) + (m * g * (L * sin a))

Dropping the common factor m and solving for x:

(16) x = [ (½ * V0²) + (g * L * sin a) ] / [ (g * sin b ) + (Crr * g * cos b) ]

= [ (½ * (1 / g) * V0²) + (L * sin a) ] / [ sin b + Crr * cos b ]

Comparing it with your equation:

(1) x = L sina / (sinb + Ur cosb),

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