A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict
1 answer:
The frictional force is 39.4 N
Explanation:
We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write
where
is the net force
m is the mass
a is the acceleration
Here we know that the box is moving with constant velocity, so its acceleration is zero:
This means that the net force is also zero:
The net force on the block is given by the applied force, forward, and the frictional force, backward:
where
is the applied force
is the frictional force
Therefore, solving for ,
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1-2) They have same surface charge density
3-4) The metallic conductor has greatest surface charge density
Explanation:
1-2)
In a conductor, the charge carriers (mainly electrons) are free to move. Therefore, as a result, they tend to move at the largest possible distance from each other, because of the repulsive force that they exert on each other.
The configuration that maximize the distance between the charge carriers for a solid sphere of metallic conductor is the one in which all the electrons are on the surface, and they are equally spaced between each other. This means that for the solid sphere of radius R, the excess charge Q will be entirely spread over the surface of the sphere.
Similarly, the excess charge Q on the hollow spherical shell (which is also made of the same conducting material) will also be spread over the surface with the charge carriers at the maximum distance from each other. Therefore, the surface charge density for both objects will be
where R is the radius of the two spheres.
3-4)
In this case, the surface charge density on the two objects is different.
In fact, on the metallic sphere (conducting) the surface charge density is (as explained in part 1):
Hoever, the second sphere is made of an insulating material. In an insulator, the charge carriers are not free to move. If the initial charge Q is spread across the all sphere (which is not hollow), this means that some of the charge will actually also be inside the sphere. So the charge deposited on the surface, Q', will be less than the total charge Q. Therefore, the surface charge density will be
And since Q' < Q, this means that , so the conducting sphere has a greatest surface charge density.