A. 1 dimensional coordinate system
As per Newton's II law we know that
![F_{net} = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma)
here we know that
![F_{net} = F_{ap} - F_f](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F_%7Bap%7D%20-%20F_f)
so here we will have
![a = \frac{F_{ap} - F_f}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF_%7Bap%7D%20-%20F_f%7D%7Bm%7D)
so here if we need to increase the acceleration we need to increase the applied force while on increasing the mass or on increasing the friction force the acceleration will decrease.
So here correct answer will be
<em>A) force on the object.</em>
Yes you're correct. For distance, SI is based on meters.
The final velocity of the truck is found as 146.969 m/s.
Explanation:
As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.
Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.
![2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}](https://tex.z-dn.net/?f=2as%20%3Dv%5E%7B2%7D%20-%20u%5E%7B2%7D%20%5C%5C%5C%5Cv%5E%7B2%7D%20%3D%202as%20%2B%20u%5E%7B2%7D)
Then, the final velocity will be
![v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%202%20%2A%203%20%2A%203600%20%2B%200%20%3D%2021600%5C%5C%20%5C%5Cv%3D%5Csqrt%7B21600%7D%3D146.969%20m%2Fs)
Thus, the final velocity of the truck is found as 146.969 m/s.
Answer:
A (2066,6 N)
Explanation:
Use the Work formula
62.000J = F . 30
62.000/30 = 2066,6 N
The amout of time it took to move the rock doesn´t matter at all.
It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.