total work done=50*4+20*2 joules
final KE= work done
A horizontal force of 50N is applied to a 2.0Kg trolly, initially at rest, and it moves a distance of 4.0m along a level, frictionless. The force then changes to 20N and acts for an aditional distance of 2.0m.
a) what is the the final kinetic energy of the trolley?
b)what is the trolley's final velocity?
2 answers
a) Work(total)= Final KE = (50*4+20*2) J
= 240J
b)Ek = 1/2 mv(final)^2 - 1/2 mv(initial)^2
v(final)^2 = Ek / 1/2 x 2
= 240 J / 1
= 240
V(final)= 15.5 m/s
= 240J
b)Ek = 1/2 mv(final)^2 - 1/2 mv(initial)^2
v(final)^2 = Ek / 1/2 x 2
= 240 J / 1
= 240
V(final)= 15.5 m/s
2 answers