A 1 kg object that stretches a spring 7.1 cm from its natural length when hangin

Question

A 1 kg object that stretches a spring 7.1 cm
from its natural length when hanging at rest
oscillates with an amplitude of 5 cm.
The acceleration of gravity is 9.81 m/s2.
a) Find the total energy of the system.
Answer in units of J.
b) Find the gravitational potential energy at
maximum downward displacement.
Answer in units of J.
c) Find the potential energy in the spring at
maximum downward displacement.
Answer in units of J.
d) What is the maximum kinetic energy of the
object?
Answer in units of J.

Explanation / Answer

Force F = kx ==>   ma = kx    ==> spring constant k = ma / x k =   (1kg)(9.81m/s^2) / (0.071 m)   = 138.03 N/m (a) total energy E = (1/2)(138.03 N/m)(0.05 m) ^2 = 0.1725 j (b) maximum gravitation potential energy U =mgA    = (1kg)( 9.8 m/s^2)(0.05) = 0.49 j (c)   maximum potential energy   of the spring U_s = (1/2)kx^2 = (1/2)(138.03N/m)(0.071m)^2 U_s = 0.347 j (d) maximum speed of the object v = sqrt (k/m (A^2 - x^2))   =0.592 m/s maximum kinetic energy of the object = (1/2)mv^2 = (1/2)(1kg)(0.592 m/s)^2 =0.1752 j

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