A 4000-kg freight car rolls along rails with negligible friction. The car is bro

Question

A 4000-kg freight car rolls along rails with negligible friction. The car is brought to rest by a combination of two coiled springs as illustrated in the figure below. Both springs are described by Hooke's law and have spring constants with k1 = 1700 N/m and k2 = 3800 N/m. After the first spring compresses a distance of 27.6 cm, the second spring acts with the first to increase the force as additional compression occurs as shown in the graph. The car comes to rest 49.5 cm after first contacting the two-spring system. Find the car's initial speed. m/s

Explanation / Answer

car’s initial kinetic energy is Ek=0.5m*v2;
final potential energy of spring system is Ep=E1+E2,

where
E1=0.5*k1*x12 is energy of spring 1,
E2=0.5*k2*x22 is energy of spring 2,

here
x1=49.5cm=0.495m, x2=49.5 cm – 27.6 cm = 0.219 m;
thus Ek=Ep
=> 0.5m*v2 = 0.5*k1*x12 +0.5*k2*x22, hence
=> v= sqrt((k1*x12 +k2*x22)/m) =
= sqrt((1700*0.4952 +3800*0.2192)/4000) =0.3869 m/s;

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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