Question

a 10.0 kg mass is at the origin. a 20.0 kg mass 0.500 m to the left of it, and a 30.0 kg mass is 1.25 m to the right of it. what is the net gravitional force on the 10.0 kg mass

Answer 1

Answer:

$4.06\cdot 10^{-8}N$ to the left

Explanation:

The gravitational force exerted between two objects is given by:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is their separation

And the force is always attractive.

Let's call

$m_1 = 10.0 kg$ the mass on which we are calculating the net force.

The mass on the left is

$m_2 = 20.0 kg$

and it is a distance of

r = 0.500 m

So the gravitational force exerted by this mass on the 10.0 kg mass is

$F_2=\frac{(6.67\cdot 10^{-11})(10.0)(20.0)}{0.500^2}=5.34\cdot 10^{-8}N$

And the direction is to the left.

The other mass is

$m_3 = 30.0 kg$

and its distance is

r = 1.25 m

to the right, so the force exerted by this other mass on the 10.0 kg mass is

$F_3=\frac{(6.67\cdot 10^{-11})(10.0)(30.0)}{1.25^2}=1.28\cdot 10^{-8}N$

And the direction is to the right.

Now, to find the net force, we just have to calculate the algebraic sum, taking into account that the two forces have different directions; taking right as positive direction, the net force is:

$F=F_3-F_2=1.28\cdot 10^{-8}N-5.34\cdot 10^{-8} N=-4.06 \cdot 10^{-8} N$

And the negative sign means the direction of the net force is to the left.

Answer 2

Answer:

-4.06 x 10^-8

Explanation:

This is the answer for acellus