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The distance travelled.
F=ma
a=25N/12kg =5.1m/s2, u=0, t=6
s=ut+1/2at2=(0*6)+1/2(5.1)(62)=33.8m
b) Workdone on the body.
W=F*s =25*33.8 =945J
c) The final kinetic energy of the body.
K.E=workdone= 945J
d) The final velocity of the body.
K.E=1/2(mv2)=945J
v={(2*945)/12}1/2= 15.6m/s.
Moraa orina answered the question on April 9, 2018 at 16:42
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