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A 13,900 kg truck is moving at 23.5 m/s on a mountain road when the brakes are applied. The brakes FAIL!! Fortunately, the driver sees a deceleration (ouch!) ramp a short distance ahead. This is an incline used by out of control trucks to dissipate their KE by converting it to PE in order to stop safely. Since the brakes have failed, you may ignore friction and air resistance.
a) What is the truck’s original kinetic energy?
b) Find the speed of the truck at a point on the ramp that is 12.5 m higher than the point where he entered the ramp.
c) How far above the point where he entered the ramp will the truck be when it comes to rest?

  • Physics -

    a) KE=mv²/2= 13900•23.5²/2 = 3.83•10⁶ J.
    b) KE= PE₁+KE₁
    KE₁ = KE-PE = mv²/2 -mgh₁ =
    =3.83•10⁶ - 13900•9.8•12.5=
    =3.83•10⁶- 1.7•10⁶ =2.13•10⁶ J.
    mv₁²/2 = KE₁
    v₁ =sqrt{ 2•KE₁/m }= sqrt{2•2.13•10⁶/13900 }=
    =17.5 m/s

    KE =PE
    mv²/2 = mgh
    h= mv²/2mg =
    =v²/2g = 23.5²/2•9.8 = 28.2 m

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