Newton's 3rd law

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Jt00

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A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. Is the rope in danger of breaking?

Answer: 565 N, yes

We have to show the work to get to the answer.
Force=mass(acceleration).
So F=50(acceleration). To get acceleration, I use 9.8 m/s2 and add it to the initial acceleration, 1 m/s2, which is 10.8 m/s2. So F=50(10.8)=540. Where did I go wrong? Dividing 565 by 50 gives me 11.3, so acceleration has to be 11.3. Subtracting 9.8 from 11.3 gives me 1.5, but wouldn't the bucket have to accelerate at just 1 m/s/s (m/s2) to be going 3 m/s after being lifted 3 m?


Sorry it's not strictly algebra, but I thought this would be the best section to post this thread in.
 
A 50.0 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. Is the rope in danger of breaking?

Answer: 565 N, yes

We have to show the work to get to the answer.
Force=mass(acceleration).
So F=50(acceleration). To get acceleration, I use 9.8 m/s2 and add it to the initial acceleration, 1 m/s2, which is 10.8 m/s2. So F=50(10.8)=540. Where did I go wrong? Dividing 565 by 50 gives me 11.3, so acceleration has to be 11.3. Subtracting 9.8 from 11.3 gives me 1.5, but wouldn't the bucket have to accelerate at just 1 m/s/s (m/s2) to be going 3 m/s after being lifted 3 m?
Click to expand...

Where does it say the initial acceleration was 1 m/s^2? It just says, "after being lifted 3.0 m, it is moving at 3.0 m/s." When it is moving at 3 m/s, this is a constant velocity; constant velocity means zero acceleration. While it is moving at constant velocity, the only forces acting on the bucket are gravity and the tension in the rope (which must be equal and opposite forces), so the tension in the rope is F = ma = (50.0 kg)(9.81 m/s^2) = 491 N.
 
Well, to accelerate to a velocity of 3 m/s in 3 m, the bucket would have to accelerate at an average rate of 1 m/s, so that at 1 m it would be going 1 m/s, at 2 m it would be going 2 m/s, and at 3 m it would be going 3 m/s. I think it is implied that the bucket continues to accelerate at this rate. But to get to the answer, there has to be an additional 1/2 m/s2 of acceleration in addition to the 1 m/s2 and gravity's 9.8 m/s2. Still, there is the possibility that the answer given was wrong.
 
Well, to accelerate to a velocity of 3 m/s in 3 m, the bucket would have to accelerate at an average rate of 1 m/s, so that at 1 m it would be going 1 m/s, at 2 m it would be going 2 m/s, and at 3 m it would be going 3 m/s. I think it is implied that the bucket continues to accelerate at this rate. But to get to the answer, there has to be an additional 1/2 m/s2 of acceleration in addition to the 1 m/s2 and gravity's 9.8 m/s2. Still, there is the possibility that the answer given was wrong.
Click to expand...

Ah, I understand now. Is there something in the problem statement that says you should assume uniform acceleration? Also, your logic is incorrect re the 1 m/s^2. Since we know initial and final velocities and the distance covered, we use the formula:

(vf)^2 = (vi)^2 + 2ax

where vf = final velocity, vi = initial velocity, and x = distance. So,

(3 m/s)^2 = (0 m/s)^2 + 2(a)(3 m)

a = 1.5 m/s^2

Hope that helps.
 
Yes, that's how it was, she explained that problem in class because almost everyone had asked her about it.
Thanks!
 
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