Question #84f65

1 Answer
MetaPhysik
Jan 11, 2018

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 energy=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

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