The only force acting on a 3.1 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.4 m/s in the positive x direction and some time later has a velocity of 6.4 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time

33.5J

Explanation:

Given:

Mass of the canister= 3.1 kg

Initial velocity of canister= v(i)= 4.4i m/s

Final velocity of canister= v(f)= 6.4j m/s

Force magnitude( xy plane)= 5 N

The magnitude of vector V'= Vxi + Vyj + Vzk

|V|= √( Vx^2 + Vy^2 + Vz^2

From Kinectic energy and work theorem.

Net work = Kinectic energy of the canister

ΔK= W

(Kf - Ki)= W

Where Kf= final Kinectic energy

= 1/2 mv^2

If we input the given values we have,

= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J

Ki= initial Kinectic energy

= 1/2 mv^2

If we substitute the given values we have

=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J

Work done by canister = (final Kinectic energy - initial energy)

= 63.5- 30

=33.5J

Hence, work done on the canister 33.5J

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.