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
You are true, Stokey. The correct choice is<em> (c)</em>.
If you think about it a little bit more, you'll realize that "side to side" is also a horizontal motion.
Anyway, the question told us that we're only talking about 2 dimensions. Well, the The measurements in the 2 dimensions are the x and the y . If there were a z measurement, that would be the 3rd dimension.
Answer:
Final velocity will be equal to 14 m/sec
Explanation:
We have given initial velocity u = 5 m/sec
Constant acceleration is given 
Time t = 6 sec
We have to find the final velocity
From first equation of motion 
, here v is final velocity, u is initial velocity , a is acceleration and t is time
So 
So equal final velocity will be equal to 14 m/sec
Mechanical advantage = output force/input force
MA = Fo/Fi
