Question

a block of mass is 3kg is kept on the top of another block of mass 5kg coefficient of friction between two block is 0.5 and 0 between 5 kg block and surface a force of 24n is applied in 5kg block than the work done by frictional force of 3kg block in t=2sec g=10m/s​

Answer 1

Answer:

block B is over block A, so normal force between the blocks will be weight of block B. =30N.

so maximum friction force that can be generated is f=μ.N=0.5×30=15N

so maximum acceleration of block B is

=

$\frac{f}{m} = \frac{15}{3} = 5m \div {s}^{2}$

so minimum acceleration of both the blocks can be

$\frac{5m}{ {s}^{2} }$

if both blocks don't want to move together.

so minimum force that required for relative motion is m.a = 8× 5 = 40 N

means force should be less then 40N, converting to (in kg wt.)(divide by 10) we get maximum force is less then 4kg wt.

so from the above options we found option B as maximum but less then 4, so option B is answer.

hope this helps you

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