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

Answer:
2156 J
Explanation:
From the question,
Work done = Combined mass of the bucket and water×height×gravity.
W = (M+m)hg............................. Equation 1
Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.
Given: M = 20 kg, m = 2 kg, h = 10 m
Constant: g = 9.8 m/s²
Substitute these value into equation 1
W = (20+2)×10×9.8
W = 22×98
W = 2156 J
Answer:
Explanation:
There is electric field between the plates whose value is given by the following expression
electric field E = V /d where V is potential between the plates and d is distance between them
E = 300 / 5 x 10⁻³
= 60 x 10³ N/c
Force on electron = q E where q is charge on the electron
F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.
Acceleration a = force / mass
a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹
= 10.55 x 10¹⁵ m / s²
For midway , distance travelled
s = 2.5 x 10⁻³ m
s = 1\2 a t²
t = 
= 
t = .474 x 10⁻¹⁸ s
For striking the plate time is calculated as follows
t =
[/tex]
t = 0.67 x 10⁻¹⁸ s