4.5 Twenty train cars with a mass of 30,000 kg each are going 45 mph when the br
ID: 1656190 • Letter: 4
Question
4.5 Twenty train cars with a mass of 30,000 kg each are going 45 mph when the brakes are applied. What average force must be applied to bring the cars to a stop over a distance of half a mile? Give your answer in both Newtons and pounds. 4.6 Two forces act on a 30 kg block, as shown in the diagram. The surface beneath the block is frictionless. a) What is the weight of the block? b) What is the normal force on the block? c) What is the net force on the block? d) What is the acceleration of the block? e) If the bloclk begins from rest, what is the speed 5 s later? 70 N 50° 100N>Explanation / Answer
(4.5) Mass of the 20 train cars, M = 20 x 30000 = 6.0 x 10^5 kg
convert the units in SI units.
45 mph = 20.1 m/s
0.5 mile = 804.7 m
use the following expression to determine the acceleration -
v^2 = u^2 + 2*a*s
=> 0 = 20.1^2 + 2*a*804.7
=> a = -20.1^2 / (2x804.7) = -0.25 m/s^2
So, the requisite force, F = M*a = 6.0x10^5x0.25 = 1.5 x 10^5 N
= 15295743 pond.
(4.6) (a) Weight of the block, W = mg = 30x9.81 = 294.3 N
(b) Normal force on the block = mg - 70*sin50 = 294.3 - 53.6 = 240.4 N
(c) Horizontal force, Fx = 100 - 70*cos50 = 55 N
Vertical force, Fy = normal force = 240.4 N
So, the net force, F = sqrt[Fx^2 + Fy^2] = sqrt[55^2 + 240.4^2] = 246.6 N
(d) Acceleration of the block = Fx / m = 55/ 30 = 1.83 m/s^2 [please note that there is no friction so we have not counted the vertical reaction]
(e) v = u + a*t = 0 + 1.83*5 = 9.15 m/s.
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