Respuesta :
Answer:
a) K_e = 0.1225 J, b) U = 1.96 J, c) v = 0.99 m / s
Explanation:
Let's use the simple harmonium movement expression
y = A cos (wt + Ф)
indicate that the amplitude is
A = 0.05 m
as the system is released, the velocity at the initial point is zero
v = dy / dt
v = - A w sin (wt + Ф)
for t = 0 s and v = 0 m/s
0 = - A w sin Ф
so Ф = 0
the expression of the movement is
y = 0.05 cos wt
The total energy of the system is
Em = ½ k A²
let's use conservation of energy
starting point. Spring if we stretch and we set the zero of our system at this point
Em₀ = K_e + U
Em₀ = 0
final point. When weight and elastic force are in balance
Em_f = K_e + U
Em_f = ½ k y² + m g (-y)
energy is conserved
Em₀ = Em_f
0 = ½ k y² + m g (-y)
k = 2mg / y
k = 2 4.00 9.8 / 0.050
k = 98 N / m
a) maximum elastic energy
K_e = ½ k A²
K_e = ½ 98 0.05²
K_e = 0.1225 J
b) the maximum gravitational energy
U = m g y
U = 4.00 9.8 0.05
U = 1.96 J
c) The maximum kinetic energy occurs when the spring is not stretched
U = K
mg h = ½ m v²
v = √2gh
v = √( 2 9.8 0.05)
v = 0.99 m / s
d) energy at any point
Em = K + U
Answer:
a) 0, b) 0, c) 3.92, d) 3.92
Explanation:
I have a similar problem to yours.
a) Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its highest point.
the answer is ((zero joules)) because we are at equilibrium on our highest point.
b) Calculate the kinetic energy of the cat when the cat is at its highest point.
the answer is ((zero joules)) because we are not moving, we are at rest.
c) Calculate the gravitational potential energy of the system relative to the lowest point of the motion when the cat is at its highest point.
from my instructor's lecture notes, we find that potential energy is found by
PE = 2mgA = 2(4kg)(9.8m/s^2)(0.05m) = 3.92 Joules
d) Calculate the sum of these three energies when the cat is at its highest point.
basically 0J + 0J + 3.92J = 3.92J
e) Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its lowest point.
same as part c)
f) Calculate the kinetic energy of the cat when the cat is at its lowest point.
same as part b)
g) Calculate the gravitational potential energy of the system relative to the lowest point of the motion when the cat is at its lowest point.
same as part a) and b)
h) Calculate the sum of these three energies when the cat is at its lowest point.
same as part d)
i) Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its equilibrium position.
from my instructor's lecture notes, the answer is:
PE(elastic) = (1/4)(3.92J) = 0.98 Joules
j) Calculate the kinetic energy of the cat when the cat is at its equilibrium position.
from my instructor's lecture notes
same as part i)
k) Calculate the gravitational potential energy of the system relative to the lowest point of the motion when the cat is at its equilibrium position.
from my instructor's lecture notes
PE = (1/2)(3.92J) = 1.96J
l) Calculate the sum of these three energies when the cat is at its equilibrium position.
the answer is: 1.96J + 0.98J + 0.98J = 3.92 J
