Question #252814

A 200kg block is pushed along a horizontal frictionless table a distance of 3meter,by a horizontal force of 12newton. Find
i)how much work is done by the force
ii)the final kinetic energy of the block
iii)the final velocity
iv)Using newton's second law,find the acceleration and then the final velocity

Expert's answer

i)

W=Fd=12⋅3=36W=Fd=12\cdot3=36 J

F - horizontal force, d - distance


ii)

kinetic energy:

E=W=36E=W=36 J


iii)

E=mv2/2E=mv^2/2

v=2E/m=2⋅36/200=0.6v=\sqrt{2E/m}=\sqrt{2\cdot36/200}=0.6 m/s


iv)

ma=Fma=F

a=F/m=12/200=0.06a=F/m=12/200=0.06 m/s2

v=at+v0v=at+v_0

d=at2/2+v0t+d0d=at^2/2+v_0t+d_0

if start velocity v0=0v_0=0 and start distance d0=0d_0=0 , then:

t=2d/at=\sqrt{2d/a}

final velocity:

v=a2d/a=0.062⋅3/0.06=0.6v=a\sqrt{2d/a}=0.06\sqrt{2\cdot3/0.06}=0.6 m/s


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