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bearhunter [10]
3 years ago
15

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. Th

e total mechanical energy is 0.12 J. What is the greatest extension of the spring from its equilibrium length? What is the maximum speed of the block?
Physics
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:

The greatest extension of the spring is \bf{0.055~m} and the maximum speed of the block is \bf{0.695~m/s}.

Explanation:

Given:

The mass of the block is, m = 0.50~kg

The spring constant of the spring is, k = 80~N/m

The mechanical energy of the block is, E = 0.12~J

When a particle is oscillating in a simple harmonic way, its total energy is given by

E = \dfrac{1}{2}m\omega^{2}a^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

where \omega is the angular velocity of the mass and a is the amplitude of its motion.

The relation between angular velocity and spring constant is given by

\omega = \sqrt{\dfrac{k}{m}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Substituting equation (2) in equation (1), we have

~~~~~~&& E = \dfrac{1}{2}ka^{2}\\&or,& a = \sqrt{\dfrac{2E}{k}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Substituting 0.12~J for E and 80~N/m for k in equation (3), we can write

a &=& \sqrt{\dfrac{2(0.12~J)}{80~N/m}}\\~~~&=& 0.055~m

The relation between the maximum velocity and the amplitude is given by

v_{m} &=& \omega a\\~~~~&=& \sqrt{\dfrac{k}{m}}a~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)

Substituting 80~N/m for k , 0.50~kg for m and 0.055~m for a in equation (4), we have

v_{m} &=& \sqrt{\dfrac{80~N/m}{0.50~kg}}(0.055~m)\\~~~&=& 0.695~m/s

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