Objects with masses of 201 kg and 616 kg are separated by 0.478 m. A 51 kg mass is placed midway between them. 0.478 m b b 201 kg 51 kg 616 kg Find the magnitude of the net gravitational force exerted by the two larger masses on the 51 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N.

2.473 * 10^(-5) N

Explanation:

The gravitational force between the first object of mass 201 kg and the object with mass 51 kg which are separated by a distance 0.239 m is:

F = (G * m1 * m2) / d²

F(1,2) = (6.672 * 10^(-11) * 201 * 51) / (0.239)²

F(1,2) = 1.197 * 10^(-5) N

Taking all the masses to be on the x axis, and the 51 kg mass as the center, we see that the 201 kg mass (on the left) pulls the 51 kg mass towards itself because it is heavier. So, we say that the force between the 51 kg mass and the 201 kg mass acts towards the left.

Therefore,

F(1,2) = -1.197 * 10^(-5) N

The gravitational force between the first object of mass 51 kg and the object with mass 616 kg which are separated by a distance 0.239 m is:

F = (G * m1 * m2) / d²

F(2,3) = (6.672 * 10^(-11) * 51 * 616) / (0.239)²

F(2,3) = 3.67 * 10^(-5) N

Taking all the masses to be on the x axis, and the 51 kg mass as the center, we see that the 616 kg mass (on the right) pulls the 51 kg mass towards itself because it is heavier. So, we say that the force between the 51 kg mass and the 616 kg mass acts towards the right.

Therefore,

F(2,3) = 3.67 * 10^(-5) N

Total force is then:

Fnet = -1.197 * 10^(-5) + 3.67 * 10^(-5)

Fnet = 2.473 * 10^(-5) N

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

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