A force in the + x-direction with magnitude F(x) = 18.0N -(0.530 N/m)x is applied to a 6.00-kg box that is sitting on the horizo
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Answer:
(a) The total energy of the object at any point in its motion is 0.0416 J
(b) The amplitude of the motion is 0.0167 m
(c) The maximum speed attained by the object during its motion is 0.577 m/s
Explanation:
Given;
mass of the toy, m = 0.25 kg
force constant of the spring, k = 300 N/m
displacement of the toy, x = 0.012 m
speed of the toy, v = 0.4 m/s
(a) The total energy of the object at any point in its motion
E = ¹/₂mv² + ¹/₂kx²
E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²
E = 0.0416 J
(b) the amplitude of the motion
E = ¹/₂KA²
(c) the maximum speed attained by the object during its motion
a) E = 0
b)
Explanation:
a)
We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:
where
E is the electric field
q is the charge contained by the Gaussian surface
is the vacuum permittivity
Here we want to find the electric field at a distance of
r = 12 cm = 0.12 m
Here we are between the inner radius and the outer radius of the shell:
However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.
This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:
q = 0
Therefore, the magnitude of the electric field is also zero:
E = 0
b)
Here we want to find the magnitude of the electric field at a distance of
r = 20 cm = 0.20 m
from the centre of the shell.
Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.
Therefore, it is given by:
where in this problem:
is the charge on the shell
is the distance from the centre of the shell
Substituting, we find: