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2 years ago

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A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until

the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?

2 answers:

Marrrta [24]2 years ago

8 0

Answer:

2

Explanation:

pulling force because of it force

kakasveta [241]2 years ago

3 0

Answer:

5.9 cm

Explanation:

f: frequency of oscillation

frequency of oscillationk: spring constant

frequency of oscillationk: spring constantm: the mass

$f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} }$

in this problem we know,

F= 1.4 Hz

m= 0.26 kg

By re-arranging the formula we get

$k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m}$

The restoring force of the spring is:

F= <em>kx</em>

where

F= 1.2 N

k= 20.1 N/m

x: the displacement of the block

$x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm$

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