A 2-kg box is pushed by a force of 4 N for 2 seconds. It has an initial velocity vo = 2 m/s to the right. NOTE: Since this problem gives the TIME ∆t traveled, FIRST look at momenta values and calculate the IMPULSE using I = F ∆t = ∆p. After the momenta values are calculated, then calculate the kinetic energy and net work values.

Respuesta :

Answer:

Kf= 36 J

W(net) = 32 J

Explanation:

Given that

m = 2 kg

F= 4 N

t= 2 s

Initial velocity ,u= 2 m/s

We know that rate of change of linear momentum is called force.

F= dP/dt

F.t = ΔP

ΔP = Pf - Pi

ΔP = m v  - m u

v= Final velocity

By putting the values

4 x 2 = 2 ( v - 2)

8 =  2 ( v - 2)

4 = v - 2

v= 6 m/s

The final kinetic energy Kf

Kf= 1/2 m v²

Kf= 0.5 x 2 x 6²

Kf= 36 J

Initial kinetic energy Ki

Ki = 1/2 m u²

Ki= 0.5 x 2 x 2²

Ki = 4 J

We know that net work is equal to the change in kinetic energy

W(net) = Kf - Ki

W(net) = 36 - 4

W(net) = 32 J

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