Question

Block A of mass m rests on the plank B of
mass 3m which is free to slide on a frictionless
horizontal surface. The coefficient of friction
between the block and plank is 0.2. If a
horizontal force of magnitude 2 mg is applied
to the plank B, the acceleration of A relative
to the plank and relative to the ground
respectively, are:​

Answer 1

Answer:

The answer is 0 and 2g/5

Explanation:

Let both the plank and the block move together.

Then acceleration of the system

a = 2mg/4m = g/2

Pseudo force on A = ma = mg/2

Maximum friction force = 0.2 mg

Since pseudo force is greater than the maximum friction force body will slip on the plank.

Acceleration of A wrt ground

a = 0.2 mg /m = 0.2 g = g/5

Acceleration of plank

a' = 2mg - 0.2 mg/3m = 0.6g

Acceleration of A wrt B

a'' = 0.6g - 0.2 g = 0.4 g = 2g/5

Answer 2

Given :-

Mass of block A = m

Mass of plank B = 3m

μ = 0.2

For mass A,

f = μ${m}_{A}$g

${m}_{A}{a}_{A}$=μ${m}_{A}$g

${a}_{A}$ = 0.2g

${a}_{A}$ = g/5

For mass B,

${F}_{net}= ({m}_{A}+{m}_{B})$a'

2mg = (4m)a'

a' = g/2

Since,

${a}_{A}$<a'

Hence,

${a}_{B}$ = Force on B/Mass of B

${a}_{B}$ = 2mg-0.2mg/3m

${a}_{B}$ = 3/5 g

Again,

${a}_{AB} = {a}_{A}-{a}_{B}$

${a}_{AB} = \dfrac{g}{5}-\dfrac{3g}{5}$

${a}_{AB} = \dfrac{-2g}{5}$

Here,

Negative sign indicates about motion in opposite direction.