Question

A crate with a mass of 120 kg glides through a space station with a speed of 5.0 m/s. An astronaut speeds it up by pushing on it from behind with a force of 220 N, continually pushing with this force through a distance of 4 m. The astronaut moves around to the front of the crate and slows the crate down by pushing backwards with a force of 190 N, backing up through a distance of 2 m. After these two maneuvers, what is the speed of the crate?

Answer 1

`v_2=5.77m/s`

Step-by-step 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:

`a_1=(F_1)/(m)`

`a_2=(F_2)/(m)`

On part 1, the initial velocity is
`v_(i1)=5m/s` and
`a_1` is applied through a distance
`d_1=4m`, so we calculate the final velocity with the formula
`v_1^2=v_(i1)^2+2a_1d_1`.

This
`v_1` will be the initial velocity of part 2, and we apply the same formula to calculate the final velocity
`v_2^2=v_(i2)^2+2a_2d_2=v_1^2+2a_2d_2=v_(i1)^2+2a_1d_1+2a_2d_2`.

Substituting our accelerations:

`v_2=\sqrt{v_(i1)^2+2a_1d_1+2a_2d_2}=\sqrt{v_(i1)^2+2(F_1)/(m)d_1+2(F_2)/(m)d_2}=\sqrt{v_(i1)^2+(2)/(m)(F_1d_1+F_2d_2)}`

Which for our values is (noticing that
`F_2` must act in the negative direction, as said):

`v_2=\sqrt{(5m/s)^2+(2)/((120Kg))((220N)(4m)+(-190N)(2m))}=5.77m/s`