A rock is thrown vertically upward with initial speed v0. 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 equal to g only at the top of the flight.
- The acceleration of the rock is always less than g.
- The speed of the rock upon return to its starting point is v0.
- The rock can attain a terminal speed greater than v0 before it returns to its starting point.
(GR8677 #01)
Solution:There is a friction force:
- acceleration is not constant → (A) and (C) FALSE
- energy is not conserved so it's initial and final speed is not the same → (D) FALSE
- frictional force slows down the object, so its speed at time t has to be less than its initial speed → (E) FALSE
Math analysis:
Friction force: Ff = − kv
The equation of motion with friction force: ma = − mg − kv.
a = − g − (kv/m)
(A) FALSE
At the top of the flight, v = 0
a = − g − (k∙0/m) = − g
(B) TRUE
Moving up: v positive
a = − g − (kv/m) = − g − c
a − g
Moving Down: v negative
a = − g − [k(−v)/m] = − g + c
a − g
(C) FALSE
No comments :
Post a Comment