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conantee
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« on: December 31, 2009, 02:34:47 AM »

1. A rock is thrown vertically upward with intial speed v_0. Assume a friction force proportional to -\vec{v}, where \vec{v} is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?
(A) The acceleration of the rock is always equals to \vec{g}
(B) The acceleration of the rock is equal to \vec{g} only at the top of the filght.
(C) The acceleration of the rock is always less than \vec{g}
(D) The speed of the rock upon return to its starting point is v_0
(E) The rock can attain a terminal speed greater than v_0 before it returns to its starting point.
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conantee
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« Reply #1 on: December 31, 2009, 02:44:17 AM »

In this problem, you don't have to solve anything or write such a equation. You just need to know the nature of force, energy and velocity of the object when it reaches the highest point.

In the problem, it is said that there exists a friction force. Hence, the total force is no longer  m \vec{g}. By Newton's law of motion, the acceleration is longer always \vec{g}. So, (A) is wrong.

Next, since there is friction, the energy dissipates in the process. There is no way the rock attain the same speed (kinetic energy), nor even greater than initial speed before returning to the starting point. So, (D) and (E) are wrong.

Now, (B) and (C) come together. Well, we know that at the highest point of motion, the velocity of object is zero. So, at that point, the friction force is zero. As if it is free fall motion, the acceleration at that point is \vec{g}. So, (B) is right, and (C) is wrong.

The correct answer is (B) ,and by stat, 67 of 100 people got the right answer.
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