Question

In the figure shown, coefficient of friction between
6 kg block and surface is zero and coefficient of
friction between blocks is 0.2. A force 5 N is applied
on 6 kg block. The acceleration of 4 kg block is
(g = 10 ms-2)
(1) 1.5 ms-2
(2) 4.0 ms-2
(3) 2.5 ms-2
(4) 0.5 ms-2​

Answer 1

Answer:

normal on 4 kg block=40 N

limiting friction between blocks=μN=0.2*40=8 N

let friction between blocks be f

assuming both blocks move together

let acceleration of both blocks be a

for 6 kg block      5-f=6a   (1)

for 4 kg block       f=4a      (2)

solving 1 and 2   a=0.5 m/s² and f = 2 N

Explanation: