Question
In Fig. 8-33, a runaway truck with failed brakes is moving downgrade at 161 km/h just before the driver steers the truck up a frictionless emergency escape ramp with an inclination of ? = 20°. The truck's mass is 1.3 x 104 kg. (a) What minimum length L must the ramp have if the truck is to stop (momentarily) along it? (Assume the truck is a particle, and justify that assumption.) What is the minimum length L if (b) the truck's mass is decreased by 15% and (c) its speed is decreased by 16%?
L = length of ramp
Explanation / Answer
Data: Mass of the truck, m = 1.3 x 10^4 kg Initial speed of the truck, v = 161 km/h = 161 * ( 5 / 18 ) m/s = 44.72 m/s Angle of the ramp, = 20 deg Let Length of the ramp = L Height of ramp, h = L sin Solution: (a) Potential energy at the top = Kinetic energy at the bottom m g h = (1/2) m v^2 2 g L sin = v^2 2 * 9.8 * L * sin 20 = 44.72^2 L = 298.36 m Ans: Length of the ramp, L = 298.36 m (b) Length is independent of truck's mass. So, the answer remains same as that of previous case. Ans: Length of the ramp, L = 298.36 m Length of the ramp, L = 298.36 m (c) New speed, v' = v - 16% of v = v - 0.16 v = 0.84 v = 0.84 * 44.72 = 37.56 m/s Potential energy at the top = Kinetic energy at the bottom m g h = (1/2) m v '^2 2 g L sin = v '^2 2 * 9.8 * L * sin 20 = 37.56^2 L = 210.5 m Ans: Length of the ramp, L = 210.5 m Potential energy at the top = Kinetic energy at the bottom m g h = (1/2) m v '^2 2 g L sin = v '^2 2 * 9.8 * L * sin 20 = 37.56^2 L = 210.5 m Ans: Length of the ramp, L = 210.5 m