AP Physics

Unit 3 - Circular Motion

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Find the net gravitational force on a 2.0 kg sphere midway between a 4.0 kg sphere and a 7.0 kg sphere that are 1.2 m apart.

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The net gravitational force on the 2.0 kg sphere will be the vector sum of the forces exerted by the 4.0 kg and 7.0 kg spheres. Since the 2.0 kg sphere is midway, each force will act along the line connecting the centers of the spheres. See the FBD below.

Step Formula Derivation Reasoning
1 rhalf=1.2,m2=0.6,m r_{\text{half}} = \frac{1.2 , \text{m}}{2} = 0.6 , \text{m} Half the distance between the two spheres.
2 F4 to 2=G4×20.62 F_{\text{4 to 2}} = G \frac{4 \times 2}{0.6^2} Gravitational force between the 4.0 kg and 2.0 kg spheres.
3 F7 to 2=G7×20.62 F_{\text{7 to 2}} = G \frac{7 \times 2}{0.6^2} Gravitational force between the 7.0 kg and 2.0 kg spheres.
4 Fnet=F7 to 2F4 to 2 F_{\text{net}} = F_{\text{7 to 2}} – F_{\text{4 to 2}} Net force is the difference between the two forces, as they are in opposite directions.

Let’s calculate the net gravitational force.

Step Formula Derivation Reasoning
4 Fnet1.11×109,N F_{\text{net}} \approx 1.11 \times 10^{-9} , \text{N} Calculated net gravitational force.

The net gravitational force on the 2.0 kg sphere, located midway between the 4.0 kg and 7.0 kg spheres (1.2 m apart), is approximately 1.11×109,Newtons \boxed{1.11 \times 10^{-9} , \text{Newtons}} . This force is directed towards the 7.0 kg sphere due to its larger mass.

 


  • The blue arrow represents the gravitational force (F42 F_{4 \to 2} ) exerted by the 4.0 kg sphere.
  • The green arrow represents the gravitational force (F72 F_{7 \to 2} ) exerted by the 7.0 kg sphere.

These forces indicate the gravitational pull exerted on the 2.0 kg sphere from each of the other two spheres. In this scenario, the net force is the vector sum of these two forces, with the direction towards the 7.0 kg sphere due to its larger mass.

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Fnet = 1.11 x 10-9