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2 years ago

A 20 kg weight is suspended from a string. What is the tension on the string

2 answers:

7 0

The weight is in equilibrium, so the two forces acting on it (tension T pulling upward, and its own weight W pulling it downward) cancel each other. By Newton's second law,

∑ F = T - W = 0

so that

T = W

Calculate the object's weight:

W = (20 kg) (9.8 m/s²) = 196 N

So the tension is T = 196 N as well.

JulsSmile [24]2 years ago

3 0

assume that tension in string a is Ta and in string b is Tb

now taking the component of tensions

Tasin30∘=Tbsin30∘...(1)

it means Ta=Tb

Tacos30∘+Tbcos30∘=mg...(2)

⇒2Tacos30∘=20×10⇒Ta=Tb=3200

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2 years ago

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And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: