A single mass m 1 = 3.6 kg hangs from a spring in a motionless elevator. The spr

Question

A single mass m1 = 3.6 kg hangs from a spring in a motionless elevator. The spring is extended x = 15 cm from its unstretched length.

1) What is the spring constant of the spring? N/m

Now, three masses m1 = 3.6 kg, m2 = 10.8 kg and m3 = 7.2 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above.

2) What is the force the top spring exerts on the top mass? N

3) What is the distance the lower spring is stretched from its equilibrium length? cm

4) Now the elevator is moving downward with a velocity of v = -3.2 m/s but accelerating upward with an acceleration of a = 4 m/s2. (Note: an upward acceleration when the elevator is moving down means the elevator is slowing down.)

What is the force the bottom spring exerts on the bottom mass? N

5) What is the distance the upper spring is extended from its unstretched length? cm

6) What is the distance the MIDDLE spring is extended from its unstretched length? cm

Explanation / Answer

a)

Since

m1g =kx

K=m1g/x =3.6*9.8/0.15

K=235.2 N/m

b)

F=(m1+m2+m3)g =(3.6+10.8+7.2)*9.8

F=211.7 N

c)

X=m3g/K =7.2*9.8/235.2

X=0.3 m= 30 cm

d)

F3=m3(g+a) =7.2(9.8+4)

F3=99.4 N

e)

Net force

F=(m1+m2+m3)(g+a) =(3.6+10.8+7.2)*(9.8+4)

F=298.1 N

X=F/K =298.1/235.2

X=126.7 cm

f)

Net force of middle spring

F=(m2+m3)(g+a) =(10.8+7.2)(9.8+4)

F=248.4 N

X=F/K =248.4/235.2

X=105.6 cm

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