Question

1. Find the force exerted by 5 kg block on floor of lift, as shown
in figure. (Take, g = 10 ms?)
5 ms-2
2 kg
5 kg
(a) 100 N (b)115 N
(c) 105 N (d)135N​

Answer 1

Answer:70 + 35

Explanation:

Normal reaction of blocks will be 5×10 + 2×10= 70

And lift is accelerated with 5m per sec^2

So it will also give an extra force of magnitude 7×5= 35N

Hence total force by block system is 105N

NOTE: lift is accelerating therefore there is an extra normal reaction it would not have been there if the lift had constant velocity.

Answer 2

Answer:

if it is for 10kg and 5 kg blocks