Answer:
[tex]v_2=5.77m/s[/tex]
Explanation:
We separate the problem in part 1 (speeding up the crate) and 2 (slowing down the crate). We take the direction the crates moves on at the beginning of the problem as the positive direction.
First we calculate the accelerations on each part using Newton's 2nd Law, which says F=ma. We will have:
[tex]a_1=\frac{F_1}{m}[/tex]
[tex]a_2=\frac{F_2}{m}[/tex]
On part 1, the initial velocity is [tex]v_{i1}=5m/s[/tex] and [tex]a_1[/tex] is applied through a distance [tex]d_1=4m[/tex], so we calculate the final velocity with the formula [tex]v_1^2=v_{i1}^2+2a_1d_1[/tex].
This [tex]v_1[/tex] will be the initial velocity of part 2, and we apply the same formula to calculate the final velocity [tex]v_2^2=v_{i2}^2+2a_2d_2=v_1^2+2a_2d_2=v_{i1}^2+2a_1d_1+2a_2d_2[/tex].
Substituting our accelerations:
[tex]v_2=\sqrt{v_{i1}^2+2a_1d_1+2a_2d_2}=\sqrt{v_{i1}^2+2\frac{F_1}{m}d_1+2\frac{F_2}{m}d_2}=\sqrt{v_{i1}^2+\frac{2}{m}(F_1d_1+F_2d_2)}[/tex]
Which for our values is (noticing that [tex]F_2[/tex] must act in the negative direction, as said):
[tex]v_2=\sqrt{(5m/s)^2+\frac{2}{(120Kg)}((220N)(4m)+(-190N)(2m))}=5.77m/s[/tex]
