A box of mass 4.0 kg hangs by a cable from the ceiling of an elevator. The elevator is moving up at a steady speed of 5.0 m/s for 8.0 m. Use g = 10 m/s2. Match the answers with questions. the final kinetic energy of the box (Kf) the change in kinetic energy of the box (ΔK)A.-20 JB.-320 JC.20 JD.50 JE.0 JF.320 JG.-50 J

Name: A box of mass 4.0 kg hangs by a cable from the ceiling of an elevator.
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: A box of mass 4.0 kg hangs by a cable from the ceiling of an elevator. The elevator is moving up at a steady speed of 5.0 m/s for 8.0 m. Use g = 10 m/s2. Match the answers with questions.

Given:- A box of mass m = 4.0 kg The elevator is moving up at a steady speed u = 5.0 m/s The height h = 8.0 m g= 10 m/s2 Find:- a) The final kinetic energy of the box (Kf) b) The change in kinetic energy of the box (ΔK)Solution:- a) We know that the velocity is given as The third equation of the motion is given as v2=u2+2ghv2=52+2×10×8v2=25+160=185v= 13.60 m/s So the final velocity is v = 13.60 m/s The final kinetic energy is given as Kf=12mv2=12×4×13.602Kf= 369.92≈370 Kf=370 J The final kinetic energy is 370 J The initial kinetic energy is given as Ki=12mu2=12×4×52Ki= 50J b) So the change in the kinetic energy is given as ∆K.E= Kf-Ki∆K.E= 370-50∆K.E = 320 J The kinetic energy is given as 320 J