Question

A runaway truck ramp is designed to slow to a stop a truck that has failed brakes. Assume that a 3 Mg truck is out of control with failed brakes and enters a runaway ramp that goes up at a constant 30 degrees above the horizontal. If the truck travels L = 400 m up the ramp before coming to a rest, what was its speed when it entered the ramp? Assume the truck rolls without friction up the ramp.

Answer 1

Vf = 62.61 m/s

Step-by-step explanation:

Since, the truck completely stops t highest point. Therefore, from conservation of energy principle:

Kinetic Energy Lost by Truck = Potential Energy Gained by Truck

(0.5)(m)(Vf² - Vi²) = mgh

(0.5)(Vf² - Vi²) = gh

where,

Vf = final velocity = 0 m/s (since, truck finally stops)

Vi = initial velocity = ?

g = acceleration due to gravity = - 9.8 m/s² (negative sign due to upward motion)

h = Height of Ramp = (Length)(Sin θ) = (400 m)(Sin 30°) = 200 m

Therefore,

(0.5)(0² m/s - Vf²) = (-9.8 m/s²)(200 m)

Vf² = (1960 m²/s²)/(0.5)

Vf = √(3920 m²/s²)

Vf = 62.61 m/s