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UGA EXAMPLE 1

UGA EXAMPLE 1. 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. 2 m. 100 kg. 75 kg. F = G (m 1 m 2 / r ²)

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UGA EXAMPLE 1

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  1. UGA EXAMPLE 1 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. 2 m 100 kg 75 kg F = G (m1m2 / r²) m1 = 100 kg m2 = 75 kg r = 2 m F = 6.67 x 10 –11Nm²/kg² (100kg * 75kg) / (2m)² = 1.25 E -7 N

  2. UGA EXAMPLE 2 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. 0.478 m 365 kg 765 kg F = G (m1m2 / r²) m1 = 365 kg m2 = 765 kg r = 0.478 m F = 6.67 x 10 –11 Nm²/s² (365kg * 765kg) / (.478m)² = 8.15 E -5 N

  3. EXAMPLE In progress Calculate the acceleration of gravity on the earth if you know its mass to be 5.98 E 24 kg. 0.478 m 365 kg 765 kg a= G (m2/ r²) m1 = 365 kg m2 = 765 kg r = 0.478 m F = 6.67 x 10 –11 Nm²/s² (365kg * 765kg) / (.478m)² = 8.15 E -5 N

  4. Calculate μ when a 6 kg box requires 3 N of force to slide at constant velocity on a horizontal surface. v= constant a= 0 m/s² Fnet = 0 FF = 3 N Fg=Fw=FN = 60 N μ= FF/ FN FN = 60 N μ= 3 N / 60 N = .05 FF= 3 N Fpull= 3 N 6 kg Fg = 6 kg (10 m/s²) = 60 N

  5. Calculate μ when a 4 kg box requires 30 N of force to slide at constant velocity on a horizontal surface. v= constant a= 0 m/s² Fnet = 0 μ= Ff / FN FF = 30 N Fg=Fw=FN = 40 N FN = 40 N μ= 30 N / 40 N = .75 FF= 30 N Fpull= 30 N 4 kg Fg = 4 kg * 10 m/s² = 40 N

  6. 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/s2? a= 0 m/s² Fnet is 0 μ= Ff / FN FN = 599 N FF = 186 N FF= 186 N Fpull= 186 N 59.9 kg Fg=Fw=FN = 599 N Fg = 59.9 kg * 10 m/s² = 599 N μ= 186 N / 599 N = 0.31

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