The acceleration of the boxes depends on the mass and weight.
we have a mass of 7 and 8 kilograms
if it took 25 N force to move box A, then you would take 25 and multiply by 8 then divide by 2.
It will leave you with 100 N.
finally take the sq rt of 100 to get 10
Answer:
See the answers below.
Explanation:
In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.
So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

Therefore we will have the following equation:
![(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]](https://tex.z-dn.net/?f=%286.5%2A9.81%2A120%29%2B%280.5%2A6.5%2A18%5E%7B2%7D%20%29%3D%286.5%2A9.81%2A60%29%2B%280.5%2A6.5%2Av_%7BB%7D%5E%7B2%7D%20%29%5C%5C3.25%2Av_%7BB%7D%5E%7B2%7D%20%3D4878.9%5C%5Cv_%7BB%7D%3D%5Csqrt%7B1501.2%7D%5C%5Cv_%7BB%7D%3D38.75%5Bm%2Fs%5D)
The kinetic energy can be easily calculated by means of the kinetic energy equation.
![KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]](https://tex.z-dn.net/?f=KE_%7BB%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7BB%7D%5E%7B2%7D%5C%5CKE_%7BB%7D%3D0.5%2A6.5%2A%2838.75%29%5E%7B2%7D%5C%5CKE_%7BB%7D%3D4878.9%5BJ%5D)
In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.
![E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]](https://tex.z-dn.net/?f=E_%7BA%7D%3DE_%7BC%7D%5C%5C6.5%2A9.81%2A120%2B%280.5%2A9.81%2A18%5E%7B2%7D%20%29%3D0.5%2A6.5%2Av_%7BC%7D%5E%7B2%7D%20%5C%5Cv_%7Bc%7D%5E%7B2%7D%20%3D%5Csqrt%7B2843.39%7D%5C%5Cv_%7Bc%7D%3D53.32%5Bm%2Fs%5D)
I THINK C BECAUSE IF IT IS A GLASS BOX HOW DID A CACTUS GET IN AND NOTHING CAN GET IN OR OUT OF THE BOX SO THERE IS NO CACTUS IN THE BOX