Chapter 15 50) A lead weight with a volume of .82X 10^-5 m^3 is lowered on a fis

Question

Chapter 15

50) A lead weight with a volume of .82X 10^-5 m^3 is lowered on a fishing line into a lake to a depth of 1.0 m. (a) What tension is required in the fishing line to give the weight an upward acceleration of 2.1 m/s^2? (b) If the initial depth of the weight is increased to 2.0 m, does the tension found in part (a) increase, decrease, or stay the same? Explain. (c) What acceleration will the weight have if the tension in the fishing line is 1.2 N? Give both direction and magnitude and please provide a step by step process.

Explanation / Answer

Look at the forces acting on the weight- there is the tension (T) up, the bouyancy force (B) up, and the weight (W) of the lead down. All these added together give you the net force on the weight. And that force is equal to the mass of the weight (m) times the accleration (a) (in this case- 2.1 m/s²). We need to know that density is equal to mass times volume and that the bouyant force on an object is equal to the weight of the fluid it displaces. Our weight displaces it's volume (V) in water, and multiplying that by the density of water (p = 1000 kg/m³) and gravity (g), we can find the bouyant force. We need to know that mass of the weight in order to find the weight of the weight (confusing?). The density of lead (j), which I looked up, is 11340 kg/m³. Multiplying this density by the volume of the lead we will find it's mass. If you have followed, here is our formula - (a) T + B - W = ma T + Vpg - (jV)g = (jV)a Solving for T, and plugging all our values in, I get- T = (11340)(.82 x 10^-5)(2.1) + (11340)(.82 x 10^-5)(9.81) - (.82 x 10^-5)(1000)(9.81) T = .195 + .912 - .080 = 1.19 N (b) The tension will remain the same! The difference in pressure is just that- a difference in pressure. It doesn't change the weight on the weight, the weight of the displaced fluid or the needed net force on the weight. (c) I wont do the calculation because we found the tension in (a) to be quite similar to 1.2 N. The acceleration will be similar too. Of course, I may have miss added or multiplied- so check me over, but I think (c) is useless

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