Question

Two blocks A and B of mass $2kg$ and $4kg$ respectively are placed one over the other as shown in the figure. A time varying horizontal force $F=2t$ is applied on the upper block as shown in figure. Here $t$ is in $second$ and $F$ is in $newton$. Draw a graph showing accelerations of A and B. The coefficient of friction between A and B is $\mu =\frac{1}{2}$ and the horizontal surface over which B is placed is smooth. $(g=10m/s^2)$

Answer 1

Limiting friction between A and B,

f${}_l$ = $\mu$m${}_a$g = $\frac{1}{2}$(2)(10) = 10N

Block B moves due to friction only, therefore maximum acceleration of B can be

a${}_{max}$ = $\frac{f_l}{m_b}$ = $\frac{10}{4}$ = 2.5 m/s${}^2$

Thus both the blocks move together with same acceleration till the common acceleration becomes 2.5 m/s², after that acceleration of B will become constant while that of A will go on increasing. To find the time when the acceleration of both the blocks becomes 2.5m/s² (or when slipping will start between A and B) we will write,

2.5 = $\frac{F}{m_a+m_b}$ = $\frac{2t}{6}$

t = 7.5 sec

Hence for t <= 7.5 S,

a${}_a$ = a${}_b$ = $\frac{t}{3}$

Thus a_A versus t or a_B versus t graph is a straight line passing through origin of slope 1/3. Hence the graph is shown above.