Answer:
a = 0.009 J
b = 0.19 m/s
c = 0.005 J and 0.004 J
Explanation:
Given that
Mass of the object, m = 0.5 kg
Spring constant of the spring, k = 20 N/m
Amplitude of the motion, A = 3 cm = 0.03 m
Displacement of the system, x = 2 cm = 0.02 m
a
Total energy of the system, E =
E = 1/2 * k * A²
E = 1/2 * 20 * 0.03²
E = 10 * 0.0009
E = 0.009 J
b
E = 1/2 * k * A² = 1/2 * m * v(max)²
1/2 * m * v(max)² = 0.009
1/2 * 0.5 * v(max)² = 0.009
v(max)² = 0.009 * 2/0.5
v(max)² = 0.018 / 0.5
v(max)² = 0.036
v(max) = √0.036
v(max) = 0.19 m/s
c
V = ±√[(k/m) * (A² - x²)]
V = ±√[(20/0.5) * (0.03² - 0.02²)]
V = ±√(40 * 0.0005)
V = ±√0.02
V = ±0.141 m/s
Kinetic Energy, K = 1/2 * m * v²
K = 1/2 * 0.5 * 0.141²
K = 1/4 * 0.02
K = 0.005 J
Potential Energy, P = 1/2 * k * x²
P = 1/2 * 20 * 0.02²
P = 10 * 0.0004
P = 0.004 J
Answer:
pi / 2 radians / s
Explanation:
One revolution = 2 pi Radians in 4 seconds
2 pi / 4 = pi/2 radians / s
Answer:
a = 3.125 [m/s^2]
Explanation:
In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.
where:
Vf = final velocity = 25 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 8 [s]
The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.
25 = 0 + a*8
a = 3.125 [m/s^2]
consider the motion along the horizontal direction :
v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s
X = horizontal displacement of the ball = 2.0 m
a = acceleration along the horizontal direction = 0 m/s²
t = time taken to land = ?
using the kinematics equation
X = v₀ t + (0.5) a t²
2.0 = 3.0 t + (0.5) (0) t²
t = 2/3
consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s
Y = vertical displacement of the ball = height of the table = h
a = acceleration along the vertical direction = 9.8 m/s²
t = time taken to land = 2/3
using the kinematics equation
Y = v₀ t + (0.5) a t²
h = 0 t + (0.5) (9.8) (2/3)²
h = 2.2 m
C 2.2 m
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.
And the distance 'd' is
Through the free-body diagram the tension components are given by
Here we can watch that,
Dividing both expression we have that,
Replacing the values,
PART B) Using the vertical component we can find the tension,
