The net force acting on the block is ~
So, the Answer in the boxes will be ~
Answer:
P = 140000 [Pa]
Explanation:
To solve this problem we must remember that pressure is defined as the relationship between Force on the area of a body.
In this particular problem, we are given the force acting on the upper surface of the block, including the force exerted by the atmospheric pressure.
P = F/A
where:
P = pressure [Pa] (units of Pascals)
F = force = 3.5*10⁴ [N]
A = area = 0.25 [m²]
P = 3.5*10⁴/0.25
P = 140000 [Pa]
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
