Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
One of the components that affect the period is gravity (the other is length). This gravity is basically the value of the effective acceleration that acts on the body due to gravity. When the elevator is over free fall, the effective gravity becomes zero. Mathematically this can be visualized as,
Since this value is zero, the period would tend to be infinite,
Therefore the frequency that is inversely proportional to the period would be defined as
In this way there is no frequency on the body which will not generate any oscillation on the body
Answer:
D. When the box is placed in an elevator accelerating upward
Explanation:
Looking at the answer choices, we know that we want to find out how the normal force varies with the motion of the box. In all cases listed in the answer choices, there are two forces acting on the box: the normal force and the force of gravity. These two act in opposite directions: the normal force, N, in the upward direction and gravity, mg, in the downward direction. Taking the upward direction to be positive, we can express the net force on the box as N - mg.
From Newton's Second Law, this is also equal to ma, where a is the acceleration of the box (again with the upward direction being positive). For answer choices (A) and (B), the net acceleration of the box is zero, so N = mg. We can see how the acceleration of the elevator (and, hence, of the box) affects the normal force. The larger the acceleration (in the positive, i.e., upward, direction), the larger the normal force is to preserve the equality: N - mg = ma, N = ma+ mg. Answer choice (D), in which the elevator is accelerating upward, results in the greatest normal force, since in that case the magnitude of the normal force is greater than gravity by the amount ma.