Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;
(a) the force constant of the spring
(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
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Answer:
K = 960 J
Explanation:
Given that,
Mass of a child = 20 kg
Mass of a sled = 10 kg
Speed of child on sled = 8 m/s
We need to find the kinetic energy of the sled with the child.
The total mass of child and the sled = 20 kg + 10 kg
= 30 kg
The formula for the kinetic energy of an object is given by :
Hence, the kinetic energy of the sled with the child is 960 J.
Utilize the formula: 
= Final Velocity (86 m/s)
= Initial Velocity (0 m/s)
a = acceleration (m/s²)
t = Time (100 seconds)
As a result,
86 m/s = 0 + (a)(100 seconds)
Using algebra, divide 86 m/s by 100 seconds:
86 m/s = 100a
a = 0.86 m/s²
Rounded to one decimal place: 0.9 m/s²
Let me know if you have any questions!
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