A Force of 6 N acts horizontally on a stationary mass of 2 kg for 4 sec. The K.E gained in Joule is...?

Question

a) 12 b) 144 c) 72 d) 48

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Answer 1 · 2018-03-28T23:38:54

This is a fairly integrated problem!

Recall,

#Deltax = nu_0t + 1/2at^2#,

#DeltaF=ma#,

#W = F*Deltax*costheta#, and by extension,

#W = Delta"KE"#

Assuming there are no other forces for clarity's sake,

#6"N" = 2"kg" * a#
#therefore a approx (3"m")/("s"^2)#

#Deltax = 1/2at^2 approx 24"m"#

Hence,

#W = 6"N" * 24"m" * 1 approx 144"J" = Delta"KE"#