Question

Question #84f65

Answer 1

`F"net"= 20/t [N]` or
`F"net"= 600/d [N]`

Where [N] is the unit Newton, t = time duration and d= distance traveled by the force.

Explanation:

You can find the unbalanced force provided you have the additional information required, i.e., either the time or distance involved in the change :

`F_"net" =ma rArr F"net" = m (v_2-v_1)/t`

`F_"net" =(2kg(20m/s-10m/s))/t =20/t(kgm/s)`

or you can use work energy principle,
`W_"net"= DeltaKrArr F_"net"d = DeltaK`
where
`W_net =` net work
`F_"net" =` unbalanced net force,
`d =` distance traveled by the force
`K=` kinetic energ`y=1/2mv^2`
`DeltaK = K_2-K_1`

Hence,
`F_"net" =( DeltaK)/F_"net" = (1/2m_2v^2-1/2mv_1^2)/d`
`F_"net" =½(4kg)((20m)^2-(10m)^2)/d = 600J/d`