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The 0.100 kg sphere in (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m from the center of the 5.00 kg mass. Assume that the only forces on the 0.100 kg sphere are the gravitational forces exerted by the other two spheres and that the 5.00 kg and 10.0 kg spheres are held in place at their initial positions.

Answer :

ajeigbeibraheem

Answer:

the speed of 0.10 sphere when it is moved to 0.20 kg  to the left is :   [tex]3.3337*10^{-5} \ m/s[/tex]

Explanation:

Using the expression of the Change in  Gravitational Potential Energy:

[tex]U= -(\frac{ Gm_1m_2 }{r_2} - \frac{ Gm_1m_2 }{r_1}) \\ \\ U=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]

So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :

[tex]U_1=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]

[tex]U_1=6.67*10^{-11}*0.1 \ kg*10 \ kg(\frac{ 1 }{0.6 \ m} - \frac{ 1 }{0.8 \ m})[/tex]

[tex]U_1 = 2.778*10^{-11} J[/tex]

the change in the gravitational potential energy  when the sphere exerts just 5 kg mass is ;

[tex]U_2=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]

[tex]U_2=6.67*10^{-11}*0.1 \ kg*5 \ kg(\frac{ 1 }{0.4 \ m} - \frac{ 1 }{0.2 \ m})[/tex]

[tex]U_2 = -8.335*10^{-11} J[/tex]

The net total change is:

[tex]U_{total } = U_1 +U_2[/tex]

[tex]U_{total} = 2.778*10^{-11} + (-8.335*10^{-11})[/tex]

[tex]U_{total} = -5.557*10^{-11}[/tex]

We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .

SO;

K.E =  [tex]5.557*10^{-11}[/tex]

[tex]\frac{1}{2}mv^2 = 5.557*10^{-11}[/tex]

[tex]v^2 = \frac{2*5.557*10^{-11} \ J}{m_1}[/tex]

[tex]v=\sqrt{ \frac{2*5.557*10^{-11} \ J}{0. 1 \kg}[/tex]

v = [tex]3.3337*10^{-5} \ m/s[/tex]

Thus, the speed of 0.10 sphere when it is moved to 0.20 kg  to the left is :   [tex]3.3337*10^{-5} \ m/s[/tex]

shubhamchouhanvrVT

The speed of 0.10 sphere when it is moved to 0.20 kg  to the left is :3.3337 *10^-5 m/s

What is speed ?

The speed of any object is defined as the movement of any object with respect to the time.

By using the expression of the Change in  Gravitational Potential Energy:

[tex]U=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex][tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]

So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :

[tex]U_1=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]

[tex]U_1=- 6.67\times 10^{-11}\times 0.1\times 10(\dfrac{1}{0.6}-\dfrac{1}{0.8})[/tex]

[tex]U_1=2.778\times 10^{-11][/tex]

The change in the gravitational potential energy  when the sphere exerts just 5 kg mass is ;

[tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]

[tex]U_2=- 6.67\times 10^{-11}\times 0.1\times 5(\dfrac{1}{0.4}-\dfrac{1}{0.2})[/tex]

[tex]U_2=-8.335\times 10^{-11}\ J[/tex]

The net total change is:

[tex]U_{total}=U_1+U_2[/tex]

[tex]U_{total}=2.778\times 10^{-11}+(-8.335\times 10^{-11})[/tex]

[tex]U_{tot\leq al}=-5.557\times 10^{-11}[/tex]

We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .

SO;

[tex]\rm KE=5.557\times 10^{-11}[/tex]

[tex]\dfrac{1}{2}mv^2=5.557\times 10^{-11}[/tex]

[tex]v^2=\dfrac{2\times 5.557\times 10^{-11}}{m}[/tex]

[tex]v=\sqrt{\dfrac{2\times 5.557\times 10^{-11}}{m}}[/tex]

[tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]

Thus, the speed of 0.10 sphere when it is moved to 0.20 kg  to the left is : [tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]

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