Question

Two blocks $A$ and $B$ of masses $5$kg and $3$kg respectively rest on a smooth horizontal surface with $B$ over $A$. The coefficient of friction between $A$ and $B$ is $0.5$. The maximum horizontal force (in kg wt.) that can be applied to $A$, so that there will be motion of $A$ and $B$ without relative slipping, is?

• A. $1.5$
• B. $2.5$
• C. $4$
• D. $5$

Answer 1

The correct option is B $2.5$

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 $f=\mu .N=0.5\times 30=15N$

so maximum acceleration of block B is $\frac{f}{m}=\frac{15}{3}=5m/s^2$

so minimum acceleration of both the blocks can be $5m/s^2$ if both blocks don't want to move together.

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

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.