Answer:
a. Yes, because the acceleration of the crate is 2.0 m/s².
Explanation:
Given
--- f
--- m
--- t
--- v
Required
Does the system support ![F=ma](https://tex.z-dn.net/?f=F%3Dma)
Yes, it does and this is shown below
The crate is initially at rest; so:
![u = 0](https://tex.z-dn.net/?f=u%20%3D%200)
Using the first equation of motion
![v = u + at](https://tex.z-dn.net/?f=v%20%3D%20u%20%2B%20at)
Substitute values for v, u and t
![3 = 0 + a*1.5](https://tex.z-dn.net/?f=3%20%3D%200%20%2B%20a%2A1.5)
![3 = 1.5a](https://tex.z-dn.net/?f=3%20%3D%201.5a)
Make a the subject
![a = 3/1.5](https://tex.z-dn.net/?f=a%20%3D%203%2F1.5)
![a = 2](https://tex.z-dn.net/?f=a%20%3D%202)
Using ![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
Substitute values for F and m
![6 = 3 * a](https://tex.z-dn.net/?f=6%20%3D%203%20%2A%20a)
Divide both sides by 3
![6/3 = 3/3 * a](https://tex.z-dn.net/?f=6%2F3%20%3D%203%2F3%20%2A%20a)
![2 = a](https://tex.z-dn.net/?f=2%20%3D%20a)
![a = 2](https://tex.z-dn.net/?f=a%20%3D%202)
In both cases:
![a = 2](https://tex.z-dn.net/?f=a%20%3D%202)
<em>Hence, option (a) is correct.</em>