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

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