Question

A box of mass 5kg is kept in side the lift. lift start from rest and moving upward with a speed 1m/s. What is the work done by normal force in first 5s

Answer 1

mass of the box is 5kg
initial velocity =0
final velocity=1m/s
time5s
f=ma
= 5kg×1-0/5
=1n

Answer 2

Given ---
Mass of box (m) = 5 kg
Initial velocity (u) = 0 m/s... (because lift starts from rest)
Final velocity (v) = 1 m/s
Time taken (t) = 5 secs

Now, we know, by first equation of motion, that,

$v = u + at$

$= > a = \frac{v - u}{t}$

Putting the values, we get,

$a = \frac{1 - 0}{5}$

$= > a = \frac{1}{5}$

$= > a = 0.2$
Hence, acceleration is 0.2 m/s²

Now,

$f = ma$

$= > f = 0.2 \times 5$

$= > f = 1$

$\green{Therefore, \: force\: exerted \:=\: 1\: N\:}$

That's your answer.

Hope it'll help.. :-)