Answer:
[tex]a=2.57\times 10^{-8}\ m/s^2[/tex]
Explanation:
Work
The work done by an external force F when moving an object by a distance X parallel to its direction is
[tex]W=F.X[/tex]
The second Newton's law gives us the net force acting on an object of mass m which is being moved at an acceleration a:
[tex]F=m.a[/tex]
Replacing the last equation into the first equation
[tex]W=m.a.X[/tex]
And solving for a
[tex]\displaystyle a=\frac{W}{m.X}[/tex]
There seems to be some kind of mistake in the data provided, we'll use it anyway
[tex]\displaystyle a=\frac{10.5}{2.72\times 10^4\times 15,000}[/tex]
[tex]\boxed{a=2.57\times 10^{-8}\ m/s^2}[/tex]
It's a very small value as compared to the dimensions of the rest of the problem.
