Question

the only force acting on a 3.5 kg canister that is moving in an xy plane has a magnitude of 7.2 n. the canister initially has a velocity of 3.2 m/s in the positive x direction, and some time later has a velocity of 4.5 m/s in the positive y direction. how much work is done on the canister by the 7.2 n force during this time?

Answer 1

17.5175J

Step-by-step explanation:

By the work-kinetic energy theorem,

`W=(1)/(2) m{v_(f) }^(2) -(1)/(2) m{v_(i) }^(2)`

where
`v_(i)` is initial velocity and
`v_(f)` is final velocity. Note that their directions make no impact on the problem.

We can simplify this to:

`W=(1)/(2) m({v_(f) }^(2) -{v_(i) }^(2))`

Now sub in our values:

`W=(1)/(2) (3.5)({4.5 }^(2) -{3.2 }^(2))`

W= (1.75)(20.25-10.24)=17.5175J