8. When a person stands on a rotating merry go round, the frictional force exerted on the person by the merry go round is
(A) greater in magnitude than the frictional force exerted on the person by the merry go round
(B) opposite in direction to the frictional force exerted on the merry go round by the person
(C) directed away from the center of the merry go round
(D) zero if the rate of rotation is constant
(E) independent of the person's mass
I have that the answer is b, but why isn't it E?
Brad, you might be dead now because you posted in2008, but it isn't independent of mass because the force of friction is the coefficient of static friction times normal force, and normal force=MASS*g.
two narrow slits 0.12 mm apart. Light of wavelength 550n illuminates the slits, causeing an interference pattern on a screen 1.0 m away. Light from each slit travels to the m = 1 maximum on the right side of the central maximum. How much farther did the light from the left slit travel then the light from the right slit? ....
where do i begin and how do i get the answer
The correct answer is (C) directed away from the center of the merry go round.
When a person stands on a rotating merry go round, they experience a centripetal force that pulls them towards the center of the rotation. In order to stay on the merry go round, the person must exert a frictional force against the surface of the merry go round. This frictional force is directed towards the center of the merry go round in order to counteract the centripetal force and allow the person to maintain their position.
The answer is not (E) independent of the person's mass because the frictional force depends on the normal force between the person and the merry go round, which is determined by the person's mass. The greater the mass of the person, the greater the normal force, and thus the greater the frictional force needed to counteract the centripetal force.
So, in summary, the frictional force exerted on the person by the merry go round is directed away from the center of the merry go round, and its magnitude depends on the person's mass.
To determine the correct answer, let's break down the question and its choices.
The question asks about the frictional force exerted on a person standing on a rotating merry-go-round. The situation involves two objects: the person and the merry-go-round. Frictional force exists between the two objects, and we need to understand its characteristics.
Now, let's consider each option:
(A) "greater in magnitude than the frictional force exerted on the person by the merry-go-round": This is not the case. According to Newton's third law of motion, the force exerted on the person by the merry-go-round is equal in magnitude and opposite in direction to the force exerted on the merry-go-round by the person. Therefore, the magnitudes of these forces are equal.
(B) "opposite in direction to the frictional force exerted on the merry-go-round by the person": This explanation is correct. When the person exerts a force on the merry-go-round, the reaction force exerted on the person by the merry-go-round (frictional force) is in the opposite direction.
(C) "directed away from the center of the merry-go-round": This statement is not entirely accurate. The frictional force between the person and the merry-go-round acts radially inward, toward the center of the circular motion.
(D) "zero if the rate of rotation is constant": This statement is incorrect. The frictional force between the person and the merry-go-round is necessary to maintain the circular motion. If this force were zero, the person would move tangentially to the circular path.
(E) "independent of the person's mass": Although the mass of the person is not directly related to the frictional force, it does not fully explain the situation. Option E does not address the direction or magnitude of the frictional force.
Based on the analysis, the correct answer is (B), as the frictional force exerted on the person by the merry-go-round is indeed opposite in direction to the frictional force exerted on the merry-go-round by the person.