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Wednesday, 10 August 2011

Physics GRE - #9

A rock is thrown vertically upward with initial speed $v_0$. Assume a friction force proportional to $-v$, where $v$ is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the following is correct?

  • The acceleration of the rock is always equal to $g$.
  • The acceleration of the rock is only equal to $g$ at the top of the flight.
  • The acceleration of the rock is always less than $g$.
  • The speed of the rock upon return is $v_0$.
  • The rock can attain a terminal speed greater than $v_0$ before it returns to its starting point.

Solution :

The second choice is the answer. 
  • Choice 1 is not true as it does not take into account the frictional force (which makes the acceleration going up greater than $g$ and the acceleration going down less than $g$).
  • Choice 2 is true as $v=0$ at the top of the flight, so the frictional force is zero and the only other force is gravity.
  • Choice 3 is not true as the acceleration going up is greater than $g$.
  • Choice 4 is not true as it does not take into account the frictional force. Friction would cause the rock to land with less kinetic energy than it started with, thus leading to a speed lower than $v_0$ upon return.
  • Choice 5 cannot be true as energy would not be conserved if speed was greater than $v_0$ (speed can only be greater than $v_0$ if friction gave the rock more kinetic energy, which is impossible since friction slows the rock by definition).

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