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3 years ago

8

The free length of the spring that is attached to the 0.3-lb slider is 3 in. If the slider is released from rest when x = 6 in.,

calculate its initial acceleration. Neglect friction.

1 answer:

hichkok12 [17]3 years ago

5 0

Answer:

a = 64 ft / s²

Explanation:

The force in a spring is given by Hooke's law

F = -k x

Let's use the initial data to calculate the spring constant

k = F / x

Reduscate to the English system

x = 3 in (1foot/12 in) =0.25 foot

k = 0.3 / 0.25

k = 1.2 lb / foot

Now we can use Newton's second law

F = ma

a = F / m

a = -k x / m

m = w / g

m = 0.3 / 32 = 0.009375

x= 6 in (1foot /12 in)= 0.5 foot

a = - 1.2 0.5 / 0.009375

a = 64 ft / s²

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b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

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$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

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Now let's begin with the answers:

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We can solve this by the conservation of energy principle:

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Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

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<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

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$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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