Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
The maximum velocity is
(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².
(b) The time when the velocity of the turtle is zero is 3.2 s.
(c) The time taken for the turtle to return to its starting point is 6.4 s.
(d) The time taken for the turtle to travel 30 cm is 0.08 s.
<h3>Initial velocity of the turtle</h3>The initial velocity of the turtle is calculated as follows;
The initial acceleration of the turtle is calculated as follows;
x(t) = 5 + 4t - 0.625t²
x(0) = 5 cm
<h3>Time when the velocity becomes zero</h3>v(t) = 4 - 1.25t
0 = 4 - 1.25t
1.25t = 4
t = 4/1.25
t = 3.2 s
<h3>Time taken to return to starting point</h3>The total distance traveled is calculated as follows
v² = u² + 2ad
0 = (4)² + 2(-1.25)d
0 = 16 - 2.5d
2.5d = 16
d = 16/2.5
d = 6.4 m
Time to travel the given distance;
d = ut + ¹/₂at²
6.4 = (4)t + ¹/₂(-1.25)t²
6.4 = 4t - 0.625t²
0.625t² - 4t + 6.4 = 0
solve the quadratic equation using formula method;
t = 3.2 s
The time travel the distance two times, = 2 x 3.2 s = 6.4 s
<h3>Time taken for the turtle to travel 30 cm</h3>d = ut + ¹/₂at²
0.3 = (4)t + ¹/₂(-1.25)t²
0.3 = 4t - 0.625t²
0.625t² - 4t + 0.3 = 0
solve the quadratic equation using formula method;
t = 0.08 s
Learn more about velocity here: brainly.com/question/6504879
Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:
Now, we can calculate the final speed of the crate at the end of 10.0 m:
For the next 10.5 meters we have frictional force:
So, the acceleration is:
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!