Asked by sweety
In Fig, a runaway truck with failed brakes is moving downgrade at 160 km/h just before the driver steers the truck up a frictionless emergency escape ramp with an inclination of è = 11°. The truck's mass is 1.5 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 11% and (c) its speed is decreased by 11%?
Answers
Answered by
drwls
The truck rises a distance of L sin 11, before momentarily stopping. At that point,
M g L sin 11 = (1/2) M Vo^2, where Vo is the initial velocity
Thus L = Vo^2/(2 g sin 11
For b, you will get the same answer, since M cancels out.
For (c), multiply the length required by (0.89)^2 = 0.792
Friction is being neglected. Often a gravel surface is used on esacape ramps to increase friction. The rotational kinetic energy of the wheels is also being neglected.
M g L sin 11 = (1/2) M Vo^2, where Vo is the initial velocity
Thus L = Vo^2/(2 g sin 11
For b, you will get the same answer, since M cancels out.
For (c), multiply the length required by (0.89)^2 = 0.792
Friction is being neglected. Often a gravel surface is used on esacape ramps to increase friction. The rotational kinetic energy of the wheels is also being neglected.
Answered by
Anonymous
If The Angle is 11degree then we can use inclined plane equation
a=gsin0-f/m....1
by using 1st law of motion we have:
vf=vi+at.....2
a=gsin0-f/m....1
by using 1st law of motion we have:
vf=vi+at.....2
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