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 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#

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