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 rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. suppose a gravel ramp slopes upward at 6.0∘ and the coefficient of rolling friction is 0.30. how long the ramp should be to stop a truck of 15000 kg having a speed of 35 m/s.

The length of the ramp required can be determined by using conservation

of energy principle.

The length of the ramp should be approximately 154.97 meters.

Reasons:

Given parameters are;

The angle of inclination of the ramp, θ = 6.0°

Coefficient of friction, μ = 0.30

Mass of the truck, m = 15,000 kg

Speed of the truck, v = 35 m/s

Required;

The length of the ramp to stop the truck

Solution:

From the law of conservation of energy, we have;

Kinetic energy = Work done against friction + Potential energy gained by the truck at height

K.E. = + P.E.

Kinetic energy of the truck, K.E. =

Therefore;

K.E. =

The kinetic energy of the truck, K.E. = 9,187,500 J

Friction force, = m·g·cos(θ)·μ

Therefore;

= 15,000 × 9.81 × cos(6) × 0.30 = 43,903.169071

Friction force, = 43,903.169071 N

Work done against friction = × d

Therefore;

Work done against friction, = 43,903.169071·d

Potential energy gained, P.E. = m·g·h

The height, h = d × sin(6.0°)

∴ P.E. = 15,000 × 9.81 × d × sin(6.0°) = 147150 × d × sin(6.0°)

Which gives;

9,187,500 J = 43,903.169071·d + 147150 × d × sin(6.0°)

The length of the ramp, d ≈ 154.97 m.

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Kinetic engery = -1/2mv^2

Work = Fd

Combine:

-1/2mv^2 - Fd = 0

-1/2mv^2 = (0.30*15000*9.81*cos(6))d = 0

Multiply both sides by 2:

v^2 = 2d(0.30*9.81*cos(6))

Solve for d:

d=v^2 / 2(0.30*9.81*cos96))

v = speed:

d = 35^2 / 2*0.30 * 9.81 * cos(6)

d = 1225 / 5.85376

d = 209.27 meters. Round answer as needed.

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