When a 0.20-kg block is suspended from a vertically hanging spring, it stretches

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

The answer is D but I keep getting 443 cm/s. I am using 1/2Kx2=1/2mv2 and my k value is 392 what am I doing wrong?

Explanation / Answer

using m*g = k*0.005 we have k = 392.3 now using energy coonserbation we have 1/2*k*( 0.15^2 - 0.05^2 ) = 1/2*m*v^2 => v = 220 cm/s (approximately)

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