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3 years ago

Mr.Z holds 200 kg above his head for 5 seconds. What is the work done on the weights?

Physics

1 answer:

Alenkasestr [34]3 years ago

8 0

m = mass held by mr. Z above his head = 200 kg

g = acceleration due to gravity = 9.8 m/s²

F = force applied by mr. Z to hold the mass

Using equilibrium of force , force equation is given as

F = mg

F = (200) (9.8)

F = 1960 N

Since the mass is not moved,

d = displacement of the mass = 0 m

we know that , work done is given as

W = F d

inserting the values

W = (1960) (0)

W = 0 J

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Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.

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Thickness, t = 50 mm, t = 0.05 m

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Using the relation, σ =P/A

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2 years ago

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