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Two students are separated in the classroom with a center to center distance of 2 m. Two students are separated in the classroom with a center to center distance of 2 m.

Two students are separated in the classroom with a center to center distance of 2 m. - PowerPoint Presentation

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Uploaded On 2018-03-17

Two students are separated in the classroom with a center to center distance of 2 m. - PPT Presentation

100 kg 75 kg F G m 1 m 2 r ² m 1 100 kg m 2 75 kg r 2 m 2 m F 667 x 10 11 Nm ²kg² 100kg 75kg 2m² 125 E 7 N UGA EXAMPLE 1 Two particles are separated in space with a center to center distance of 0478 m The mass of A is 365 kg and ID: 654388

net 365 478 186 365 net 186 478 765 mass calculate 599 force center kgm2 box slide kgr pull

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Slide1

Two students are separated in the classroom with a center to center distance of 2 m. The mass of A is 100 kg and the mass of B is 75 kg. Find the magnitude and direction of the net gravitational force acting on the students.

100 kg

75 kg

F = G (m

1

m2 / r²)m1 = 100 kgm2 = 75 kgr = 2 m

2 m

F = 6.67 x 10

–11 Nm²/kg² (100kg * 75kg) / (2m)²

= 1.25 E -7 N

UGA EXAMPLE 1Slide2

Two particles are separated in space with a center to center distance of 0.478 m. The mass of A is 365 kg and the mass of B is 765 kg. Find the magnitude and direction of the net gravitational force acting on the particles.

365 kg

765 kg

F = G (m

1

m2 / r²)m1 = 365 kgm2 = 765 kgr = 0.478 m

0.478 m

F = 6.67 x 10

–11 Nm²/s² (365kg * 765kg) / (.478m)²

= 8.15 E -5 N

UGA EXAMPLE 2Slide3

Calculate the acceleration of gravity on the earth if you know its mass to be 5.98 E 24 kg.

365 kg

765 kg

a= G (m

2

/ r²)m1 = 365 kgm2 = 765 kgr = 0.478 m

0.478 m

F = 6.67 x 10

–11 Nm²/s² (365kg * 765kg) / (.478m)²

= 8.15 E -5 NEXAMPLE In progressSlide4

Calculate μ when a 6 kg box requires 3 N of force to slide at

constant velocity on a

horizontal surface.

6 kg

F

pull= 3 NFg = 6 kg (10 m/s

²) = 60

N

FN = 60 N

v= constanta= 0 m/s

²

F

net

= 0

μ= F

F

/

F

N

F

F

= 3 N

F

g

=

F

w

=F

N

=

60

N

μ= 3 N

/

60

N

=

.05

F

F

=

3 NSlide5

Calculate μ when a 4 kg box requires 30 N of force to slide at

constant velocity on a

horizontal surface.

4 kg

F

pull= 30 NFg = 4 kg * 10

m/s² = 40 N

F

N = 40 N

v= constanta= 0 m/s

²

F

net

= 0

μ= F

f

/ F

N

F

F

= 30 N

F

g

=

F

w

=F

N

=

40

N

μ= 30 N /

40

N

=

.

75

F

F

=

30 NSlide6

Calculate μ when it takes

186 N to slide a 59.9 kg box across a horizontal slushy ice rink at an acceleration of 0 m/s

2?

59.9 kg

F

pull= 186 N

Fg = 59.9 kg * 10 m/s

² = 599 N

F

N = 599 N

a= 0 m/s

²

F

net

is 0

μ= F

f

/ F

N

F

F

= 186 N

μ= 186 N /

599

N

=

0.31

F

F

=

186 N

F

g

=

F

w

=F

N

=

599

N