A student processes a design for an automobile crash barrier in which a 1350-kg
ID: 1293994 • Letter: A
Question
A student processes a design for an automobile crash barrier in which a 1350-kg sport utility vehicle moving at 25.0 m/s crashes into a spring of negligible mass that slows it to stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater that 5.00 kg. Part A Find the required spring constant k. In your calculation, disregard or crumpling the vehicle and the friction between the vehicle and the ground. Part B Find the distance the spring will compress in slowing the vehicle to a stop.Explanation / Answer
since acceleration is 5g i.e., 5 x 9.8 = 49m/s^2.
using v^2 - u^2 = 2as
we get s = 6.38 m, (take v = 0, u = 25m/s and a = 49 m/s^2)
therefore, distance in which the vehicle comes to rest will be the compression of the spring.
therefore, the compression x = 6.38m.
then, kinetic energy of vehicle is equal to compression of the spring,
1/2 mv^2 = 1/2 k x^2
k = (mv^2)/x^2
k = (1350 x 25 x 25)/ (6.38^2)
so, finally we get
k = 20728.72 N/m.
disadvantages:
friction between movable parts
stability of the system
making a spring of high spring constant.
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