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

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

1 Answer
Mar 28, 2018

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"