Mass attached to a rope (simple)

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A 1000 kg mass is lowered by a rope, initially moving at 10 m/s and decelerating at 1 m/s². The tension in the rope must account for both the weight of the mass and the upward acceleration due to deceleration. The correct equation is T = mg + ma, leading to a tension of 11,000 N, not 9,000 N as initially calculated. The key point is that while the mass moves downward, its upward acceleration increases the tension in the rope. Understanding the direction of acceleration is crucial for solving the problem accurately.
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1. A 1000kg mass is lowered by a rope. If the mass is initially moving at 10m/s, and is decelerating at 1m/s2, what is the initial tension T in the rope?



2. T = mg



3. So I tried setting up T = mg, and then since it is decelerating the tension must be less than what it would be if it was in static equilibrium. Then I added ma to side T.

ma + T = mg

Ended up getting T = 9000N, but the answer is 11,000N.


What did I do wrong? I thought my thinking was correct. I thought about a mass in an elevator which would weigh less if elevator moved downward and weigh more if elevator moved up.
 
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saltyload said:
I thought about a mass in an elevator which would weigh less if elevator moved downward and weigh more if elevator moved up.
If the elevator accelerates downward, the mass would 'weigh' less. Acceleration, not movement, is the key.

So what's the direction of the acceleration of the 1000 kg mass? (Read carefully.)
 
Doc Al said:
If the elevator accelerates downward, the mass would 'weigh' less. Acceleration, not movement, is the key.

So what's the direction of the acceleration of the 1000 kg mass? (Read carefully.)

AH! the mass is moving downward, but since its decelerating, its accelerating upwards, which increases the tension of the rope!
 
Exactly! :wink:
 
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