Question

3. (2.5 Pts.) In the Atwood machine shown in figure below m? = 10 Kg and m? = 20 Kg (masses of pulley and rope are both negligible). How much is a?? (the acceleration of m?)? Consider the upward direction to be the positive y-axis. m? m? A) -19.6 m/s² B) 19.6 m/s² C) -9.8 m/s² D) 9.8 m/s² E) None of the above

          3. (2.5 Pts.) In the Atwood machine shown in figure below m? = 10 Kg and m? = 20 Kg (masses of pulley and rope are both negligible). How much is a?? (the acceleration of m?)? Consider the upward direction to be the positive y-axis.
m?
m?
A) -19.6 m/s²
B) 19.6 m/s²
C) -9.8 m/s²
D) 9.8 m/s²
E) None of the above

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3. (2.5 Pts.) In the Atwood machine shown in figure below m? = 10 Kg and m? = 20 Kg (masses of pulley and rope are both negligible). How much is a?? (the acceleration of m?)? Consider the upward direction to be the positive y-axis.
m?
m?
A) -19.6 m/s²
B) 19.6 m/s²
C) -9.8 m/s²
D) 9.8 m/s²
E) None of the above

Added by Antonia M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Timothy James

03/13/2023

Step 1

First, we need to find the net force acting on the system. Since m1 is going upward and m2 is going downward, the net force will be the difference between the gravitational forces acting on both masses: F_net = m2 * g - m1 * g. Show more…

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In the Atwood machine shown in figure below m1 = 10 Kg and m2 = 20 Kg (masses of pulley and rope are both negligible). How much is a₁ₘ (the acceleration of m1)? Consider the upward direction to be the positive y-axis. A) -19.6 m/s² B) 19.6 m/s² C) -9.8 m/s² D) 9.8 m/s² E) None of the above

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