Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a
runaway 60,000 kg truck is moving at 27 m/s, how much work must be done to stop the truck?

Answer :

elcharly64

Answer:

[tex]W=-21,870,000\ J[/tex]

Explanation:

Work and Kinetic Energy

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

[tex]W=\Delta k=k_1-k_0[/tex]

Since

[tex]\displaystyle k=\frac{mv^2}{2}[/tex]

We have

[tex]\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}[/tex]

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

[tex]\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}[/tex]

[tex]W=0-21870000\ J[/tex]

[tex]\boxed{W=-21,870,000\ J}[/tex]

dsdrajlin

The work done to stop a 60,000-kg truck moving at 27 m/s is -2.2 × 10⁷ J.

What does the work-energy principle state?

The principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

We can calculate the work done to stop the truck using the work-energy principle.

W = ΔK

W = 1/2 × m × v² - 1/2 × m × u²

W = 1/2 × 60,000 kg × (0 m/s)² - 1/2 × 60,000 kg × (27 m/s)²

W = -2.2 × 10⁷ J

where

  • W is the work.
  • K is the kinetic energy.
  • m is the mass.
  • v is the final speed.
  • u is the initial speed.

The work done to stop a 60,000-kg truck moving at 27 m/s is -2.2 × 10⁷ J.

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