Question

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equilibrium and released from rest to slide on a horizontal, frictionless table. Calculate the maximum speed of the object.
Find the locations of the object when its velocity is one-third of the maximum speed. Treat the equilibrium position as zero, positions to the right as positive, and positions to the left as negative.

Answer 1

Answer:

$v = 1.14 m/s$

Part b)

$x = \pm 5.89 cm$

Explanation:

Part a)

As we know that the maximum speed of the spring block system is given as

$v_{max} = A\omega$

so we have

$A = 6.25 cm$

$\omega = \sqrt{\frac{k}{m}}$

so we have

$\omega = \sqrt{\frac{74.5}{0.225}}$

$\omega = 18.2 rad/s$

now we have

$v = 0.0625 \times 18.2$

$v = 1.14 m/s$

Part b)

speed of the object at a distance x from its mean position is given as

$v = \omega\sqrt{A^2 - x^2}$

$\frac{A\omega}{3} = \omega\sqrt{A^2 - x^2}$

$\frac{A^2}{9} = A^2 - x^2$

$x = \pm\sqrt{\frac{8A^2}{9}}$

$x = \pm\sqrt{\frac{8(6.25)^2}{9}}$

$x = \pm 5.89 cm$

Answer 2

(a) The maximum speed of the object is 1.14 m/s.

(b) The locations of the object when its velocity is one-third of the maximum speed is 5.9 cm.

Maximum speed of the object

The maximum speed of the object is determined using principle of conservation of energy.

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{74.5}{0.225} }\\\\\omega = 18.2 \ rad/s$

Maximum speed = ωr = 18.2 x 0.0625 = 1.14 m/s

Location of the object

The speed of the object at a distance x from its mean position is given as;

$v = \omega \sqrt{A^2 - x^2} \\\\\frac{A\omega }{3} = \omega \sqrt{A^2 - x^2} \\\\\frac{A^2}{9} = A^2 - x^2\\\\x^2 = \frac{8A}{9} \\\\x = \sqrt{\frac{8A}{9}} \\\\x = \sqrt{\frac{8 \times 6.25^2}{9} } \\\\x = 5.9 \ cm$

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