8. A 300g mass hangs at the end of a string. A second-string hang from the bottom of that mass and supports a 900g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700m/s2. (b) Find the tension in each string when the masses are accelerating downward at 0.700m/s2.

Question

Carl Bates

Physics

24 June, 21:15

8. A 300g mass hangs at the end of a string. A second-string hang from the bottom of that mass and supports a 900g mass. (a) Find the tension in each string when the masses are accelerating upward at 0.700m/s2. (b) Find the tension in each string when the masses are accelerating downward at 0.700m/s2.

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Answers (1)

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Teagan Good · Answer 1

We are looking for:

Tension A=?

Tension B=?

Then convert grams to kilograms

1kg = 1000g

300g = 0.3 kg

900g = 0.9kg

Given:

mA = 0.3kg

mB = 0.9kg

how? W=mg

g = 9.8 m/s^2

ay = 0.700m/s^2

WA = 2.94N

WB = 8.82N

For A [Tension A-Weight A - Tension B=mAay]

TA - 2.94 - TB=0.3kg (0.700m/s^2)

TA - 2.94 - TB = 0.21 N

TA - TB = 0.21N + 2.94 N

TA - TB = 3.15 N

* refer TB from below*

TA - TB = 3.15N

TA - 9.45N = 3.15N

TA = 3.15N + 9.45N

TA = 12.6N

For B F = ma

Summation of forces along y-axis = mBay

-8.82 N + TB = 0.9 kg (0.700 m/s^2)

-8.82N + TB = 0.63 N

TB = 0.63 N + 8.82 N

TB = 9.45 N