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Ivan
3 years ago
12

A heavy ball with a weight of 100 N is hung from the ceiling of a lecture hall on a 4.3-m-long rope. The ball is pulled to one s

ide and released to swing as a pendulum, reaching a speed of 5.0 m/s as it passes through the lowest point. Part A What is the tension in the rope at that point?
Engineering
1 answer:
Vlada [557]3 years ago
6 0

Answer: T= 159.3N

Explanation:

The total Tension on the rope at this point=

Sum of the weight of the ball + the centripetal force on the ball

T= W+ mv^2/r

Where T is tension, W is weight of the ball, m is mass of the ball, r is radius of rotation, v is the velocity of rotation. g= acceleration due to gravity

W= 100N

m= W/g= 100/9.8 = 10.2kg

v= 5m/s

r= 4.3m

Therefore,

T= 100 + 10.2×5^2/4.3

T= 100 + 59.3

T= 159.3N

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Explanation:

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From steam table

<u> At 10 MPa and 500 C:</u>

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s_g= 7.3  KJ/Kg-K                ,s_f=1.3   KJ/Kg-K

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