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Veseljchak [2.6K]

2 years ago

7

Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If

the distance is 140 cm when the mass is 21 kg, what is the distance when the mass is 5 kg?

1 answer:

trasher [3.6K]2 years ago

8 0

Answer:

$d_{2}$ = 33.33 cm

Explanation:

Given:

When mass, $m_{1}$ =21 kg

distance travelled is $d_{1}$ = 140 cm

When mass, $m_{2}$ =5 kg

distance travelled is $d_{2}$ = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

$\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}$

$\frac{140}{21}=\frac{d_{2}}{5}$

$d_{2}$ = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

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Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

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