Answer:
(a) The total energy of the object at any point in its motion is 0.0416 J
(b) The amplitude of the motion is 0.0167 m
(c) The maximum speed attained by the object during its motion is 0.577 m/s
Explanation:
Given;
mass of the toy, m = 0.25 kg
force constant of the spring, k = 300 N/m
displacement of the toy, x = 0.012 m
speed of the toy, v = 0.4 m/s
(a) The total energy of the object at any point in its motion
E = ¹/₂mv² + ¹/₂kx²
E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²
E = 0.0416 J
(b) the amplitude of the motion
E = ¹/₂KA²
[tex]A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m[/tex]
(c) the maximum speed attained by the object during its motion
[tex]E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s[/tex]
a. The total energy of the toy at any point in its motion is 0.0416 Joules.
b. The amplitude of the motion is equal to 0.0167 meter.
c. The maximum speed attained by the toy during its motion is 0.577 m/s.
Given the following data:
a. To find the total energy of the toy at any point in its motion:
Mathematically, the total energy of an object undergoing simple harmonic motion (SHM) is given by:
[tex]E = \frac{1}{2} MV^2 + \frac{1}{2} kx^2[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]E = \frac{1}{2} \times 0.25 \times 0.400^2 + \frac{1}{2} \times 300 \times 0.0120^2\\\\E = 0.125 \times 0.16 + 150 \times 0.000144\\\\E=0.02+0.0216[/tex]
E = 0.0416 Joules.
b. To find the amplitude of the motion, we would use this formula:
[tex]A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2 \times 0.0416}{300} } \\\\A = \sqrt{2.77 \times 10^{-4}}[/tex]
A = 0.0167 meter.
c. To find the maximum speed attained by the object during its motion:
[tex]V_{max} = \sqrt{\frac{2E}{M} } \\\\V_{max} = \sqrt{\frac{2 \times 0.0416}{0.25} } \\\\V_{max} = \sqrt{2.77 \times 10^{-4}}[/tex]
Maximum speed = 0.577 m/s
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