Question

hi
can u help me, please

A 2Kg object is moving at 5m/s. A 8N force is applied in the direction of motion and then removed after the object has traveled an additional 2s. Calculate the work done on it.

Answer 1

The answer is 144 J . How can I get it ????

Answer 2

The Work equal
\[W(F)_{a-b}=F.d.\cos \alpha \]
F is the force (N)
d is the distance
cos alpha
if Vertically you dont use the cos alpha

Answer 3

This formula is used only when the force is constant !

Answer 4

of course
but i dont know why your force is not constant

Answer 5

i think the key to solve this is to consider that

\[W = \Delta K = \frac{ 1 }{ 2 } m (v_{f}^{2} - v_{i}^{2} )\]

but i dont know how can i get the final velocity ?!!!

Answer 6

I'm sorry I'm not at this level
...

Answer 7

ok, thanks

Answer 8

u can do this in ur method using work-energy theorem or using method by @AntarAzri
if u do by ur method,find the final velocity for the time t=2s using v=u+at
or by the other:
u should not that work is done only when a force acts onb a body so initially the body moved with CONSTANT velocity=== no force
so find the distance it moves in the presence of a force
s=ut+1/2at^2
then done