At t = 4.0 s, the elevator door opens and remains open for 5.0 s. The student immediately exerts a larger constant force on the box and the front of the box reaches the elevator door just as it start to close.

() Calculate the magnitude of the new force that the student exerts.

Explanation:

Given,

mass of block = 30 kg

initial speed = 1.2 m/s

coefficient of friction = 0.20

time for elevator close and opening = 5.0 sec

for t= 0 sec to t= 4.0 sec

acceleration = 0

speed = 1.2 m/s

distance x= vt

at t=4 distance is x=4.8m

for t= 4.0 sec to next 5 second or t= 9.0 sec

the remaining distance for cover is

x= 16-4.8

x=11.2 m

using motion equation

x=vt+1/2at^2

11.2=1.2*5+1/2 0.5aa=0.416 m/s^2

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2