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dusya [7]
2 years ago
6

A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 N and 15

10 N, respectively. The acceleration of the cable is 0.620 m/s2, upward. What is the tension in the cable(a) below the worker and(b) above the worker?
Physics
1 answer:
Alex777 [14]2 years ago
4 0

Answer:(a)2631.46 N

(b)1605.51 N

Explanation:

Assuming there is a cable between crate and worker

Given

weight of worker W_w=965 N

Weight of crate W_c=1510 N

mass of worker m_w=\frac{965}{g}=98.46 kg

mass of crate m_c=\frac{1510}{g}=154.08 kg

acceleration of system a=0.620 m/s^2

Tension in the wire carrying Worker

T-(m_w+m_c)g=(m_w+m_c)a

T=(m_w+m_c)(g+a)

T=(252.54)(9.8+0.62)

T=2631.46 N

(b)Tension in the string above

T_2-m_c\cdot=m_ca

T_2=m_c(g+a)

T_2=154.08(9.8+0.62)

T_2=1605.51 N

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Answer:

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Hi there!

a) The average velocity (a.v) is calculated as the displacement divided by the time it took to do such a displacement.

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The magnitude of your average velocity during the 121 s is 8.61 m/s.

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Using trigonometry, we can calculate the value of the angle θ:

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