Please show picture of diagrams
Make sure the triple beam balance is at 0 before you begin.
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

<h2>
Power is 11 W</h2>
Explanation:
Power = Work ÷ Time
Work = Force x Displacement
Force = 22 N
Displacement = 3 m
Time = 6 seconds
Substituting
Work = Force x Displacement
Work = 22 x 3 = 66 J
Power = Work ÷ Time
Power = 66 ÷ 6
Power = 11 W
Power is 11 W
Electrostatic forces between charges depend on the product of
the sizes of the charges, and the distance between them.
We should also mention the item about whether the charges are
both the same sign or opposite signs. That determines whether
the forces will pull them together or push them apart, which is a
pretty significant item.