the kinetic energy = (1/2) m v^2
initially the speed is 2
so the initial Ke = (1/2)(10)(4) = 20 Joules
A 10 kilogram object subjected to a 20 Newton force moves across a horizontal frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0 meters/second. When the force is removed the object is traveling at 6.0 meters per second. The initial kinetic energy is 20 J?
2 answers
Now I suspect you will be asked what the final Ke is
(1/2)(10)(36) = 180 Joules
the increase in Ke = 180 - 20 = 160 Joules = work done by force
= Force*distance
so distance = 160/20 = 8 meters
now for time
Force * time = change in momentum
20 t = 10(6-2)
t = 40/20 = 2 seconds
(1/2)(10)(36) = 180 Joules
the increase in Ke = 180 - 20 = 160 Joules = work done by force
= Force*distance
so distance = 160/20 = 8 meters
now for time
Force * time = change in momentum
20 t = 10(6-2)
t = 40/20 = 2 seconds
1 answer