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Two blocks are connected by a rope, as shown above. The masses of the blocks are [mass of upper block] and [mass of lower block]. An upward applied force of magnitude [force] acts on the upper block. Which of the following predictions about the acceleration of the two-block system is correct?

1) The acceleration is downward with a magnitude of [acceleration1].
2) The acceleration is downward with a magnitude of g.
3) The acceleration is downward with a magnitude of [acceleration2].
4) The acceleration is downward with a magnitude of 2g.
5) The acceleration is upward with a magnitude of [acceleration3].
6) The acceleration is upward with a magnitude of g.
7) The acceleration is upward with a magnitude of [acceleration4].

1 Answer

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Final answer:

To determine the acceleration of the two-block system, one must subtract the gravitational weight of both blocks from the upward applied force, and divide the net force by the total mass of the system to obtain the net acceleration, aligning with Newton's second law of motion.

Step-by-step explanation:

To predict the acceleration of the two-block system, we can apply Newton's second law and the concept of acceleration due to gravity (g). Since the acceleration due to gravity is 9.8 m/s² downward, we must consider the force and direction of the applied force in relation to the blocks' weights (force of gravity acting on them). Assuming the upward applied force is sufficient to overcome the weight of the lower block but not the combined weights of both blocks, we can ascertain a net upward force affecting the system.

Next, we calculate the net force by subtracting the weight of the two blocks (considering the mass of each block and acceleration due to gravity) from the upward applied force. The net force divided by the total mass of the system will give us the net acceleration of the system. If the net force is upward, then according to Newton's second law of motion, the acceleration will also be upward. If it is downward, the acceleration will be downward.

User Vadim Iarovikov
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