A block of mass m= 10.0 kg is attached to the end of an ideal spring. Due to the

Question

A block of mass m= 10.0 kg is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance h= 7.00 cm from its equilibrium length. (Figure 1) The spring has an unknown spring constant k. Take the acceleration due to gravity to be g = 9.81 m/s2 .

Part A What is the spring constant k?

Part B Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency of the block's oscillations about its equilibrium position. Express your answer in radians per second

Explanation / Answer

Part A

Hooke’s Law gives the formula F = -kY

Since the spring is in equilibrium (not moving) when stretched 0.05m:

F = -kY
mg = -k(Y – Y0)
(10 * 9.81) = -k(0 – 0.07)
98.1 = 0.07k
k = 11401.42N/m

k = 1401N/m (2 significant figures)

Part B

The angular frequency is given by the formula:

= sqrt(k/m)
= sqrt(1401 / 10)

= 11.83 radians/s

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