Now, three masses m1 = 4.7 kg, m2 = 14.1 kg and m3 = 9.4 kg hang from three iden
ID: 2167449 • Letter: N
Question
Now, three masses m1 = 4.7 kg, m2 = 14.1 kg and m3 = 9.4 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above.What is the force the top spring exerts on the top mass?
3) What is the distance the lower spring is stretched from its equilibrium length?
4) Now the elevator is moving downward with a velocity of v = -2.5 m/s but accelerating upward with an acceleration of a = 4.4 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?
Explanation / Answer
let... e,g... (a) By F = - kx =>k = F/x = 12/5 = 2.4 N/m [ignore the -ve sign as it indicates the direction of the force] (b) By PE(spring) = 1/2kx^2 = 1/2 x 2.4 x (1.5)^2 = 2.7 J (c) By PE(spring) = KE(cart) at the spring passes through its equilibrium position =>KE(cart) = 2.7 J =>By KE = 1/2mv^2 =>2.7 = 1/2 x 0.5 x v^2 =>v = sqrt[10.8] = 3.29 m/s
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