When a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from its original length of 0.040 m to 0.045 m. The same block is attached to the same spring and placed on a horizontal, frictionless surface as shown. The block is then pulled so that the spring stretches to a total length of 0.15 m. The block is released at time t = 0 s and undergoes simple harmonic motion.
-What is the speed of the block each time the spring is 5.0 cm long?
A)zero cm/s
B)422 cm/s
C)344 cm/s
D)220 cm/s
E)160 cm/s

Question

When a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from its original length of 0.040 m to 0.045 m. The same block is attached to the same spring and placed on a horizontal, frictionless surface as shown. The block is then pulled so that the spring stretches to a total length of 0.15 m. The block is released at time t = 0 s and undergoes simple harmonic motion. 
-What is the speed of the block each time the spring is 5.0 cm long?
A)zero cm/s 
B)422 cm/s 
C)344 cm/s 
D)220 cm/s 
E)160 cm/s

Quizplus · Accepted Answer

First, we can use Hooke's law to find the spring constant:
F = kx
mg = kx
k = mg/x = (0.20 kg)(9.81 m/s^2)/(0.045 m - 0.040 m) ≈ 98.1 N/m

Next, we can use energy conservation to find the speed of the block when the spring is 5.0 cm long:
PE at equilibrium = KE at max compression
(1/2)kx^2 = (1/2)mv_max^2
(1/2)(98.1 N/m)(0.05 m)^2 = (1/2)(0.20 kg)v_max^2
v_max = √([(98.1 N/m)(0.05 m)^2]/(0.20 kg)) ≈ 220 cm/s

Therefore, the answer is choice D, 220 cm/s.

When a 020-Kg Block Is Suspended from a Vertically Hanging Spring, It