The force exerted on a car by a prototype crash barrier as the barrier crushes i

Question

The force exerted on a car by a prototype crash barrier as the barrier crushes is F = - (1000 + 10,000s) lb, where s is the distance in feet from the initial contact. Suppose you want to design the barrier so that it can stop a 5000-lb car traveling at 80 mi/hr.

a) What is the necessary effective length of the barrier? That is, what is the distance required for the barrier to bring the car to a stop?

b) What is the rate at which power is transferred from the car when it first contacts the barrier?

Explanation / Answer

The Amount of wor is given by

U = integral ( - (1000 + 10,000s)) with limits from 0 to sf , where sf is length of barrier

80 mi/hr = 35.76 m/s

U = -1000sf-5000sf^2 = -1/2 mv^2 = -0.5*5000*35.76 m/s

1000sf+5000sf^2 -89400 =0

sf = 4.129 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.