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= 10.0 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 . What is the spring constant k? Express your answer in newtons per meter. 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

spring force = weight

kh = mg

k (0.10) = 10 x 9.8

k = 980 N/m

angular frequency is given as

w = sqrt(k/m) = sqrt(980/10) = 9.89 rad/s

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