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Tom [10]
4 years ago
15

A 60kg student is 5m from a 120kg door, how much gravitational forces act between the two?

Physics
1 answer:
slavikrds [6]4 years ago
7 0

Gravitational force = G · m₁ · m₂ / r²

' G ' is the 'universal gravitational constant, 6.673 x 10⁻¹¹ m³/kg-s²

m₁ = 60 kg

m₂ = 120 kg

r = 5 m

F = (6.673 x 10⁻¹¹ m³/kg-sec²) · (60 kg) · (120 kg) / (5m)²

F = (6.673 x 10⁻¹¹ · 60 · 120 / 25) · (m³ · kg · kg / kg · s² · m²)

F = 1.922 x 10⁻⁸ (kg · m / s²)

<em>F = 1.922 x 10⁻⁸ Newton</em>  (about 0.0000000692 ounce of force)

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As far as I'm concerned, this is a bogus question, or at least a severely corrupted one.

The three numbers given can NOT all be true on Earth.

-- It rolled off the table at 7.6 m/s .  By golly, there you are!  Its initial horizontal velocity is 7.6 m/s, and it has no vertical velocity until it leaves the table.

-- There are no horizontal forces that we're aware of acting on the object.  So it maintains the same horizontal velocity for the rest of the story.  It's 10.5m away from the table in (10.5 m) / (7.8 m/s) = 1.35 second .

-- Vertically, it's just an object dropped from 17.6m off the floor.  Shockingly, the distance it falls in time 'T' is (1/2 g) T².  In 1.35 second, that's 8.88 meters ! . . . only about halfway to the floor !

-- In order to fall 17.6 m to the floor, it would need 1.89 seconds.  In <u>that</u> length of time, however, it would travel (7.8 m/s) x (1.89 s) = 14.78 m away from the base of the table.

So you see, either . . .

-- the table is NOT 17.6m tall, or

-- the object does NOT roll off of the table at 7.8 m/s, or

-- it does NOT land 10.5 m away from the base of the table.

OR . . .

-- the table is not on Earth, and gravity is not 9.8 m/s² !

We often see questions posted on Brainly with not enough given information, OR with some information given that's not needed because it's not involved the answer.  

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(Unless the whole story takes place on a mystery planet that is not Earth.  Which I'm not going to take the time and effort right now to figure out what the acceleration of gravity has to be in order to make all of the given information compatible.)

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