A 2kg block is placed over a 5kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.10. Find the acceleration of the two blocks if a horizontal force of 14N is applied to the upper block (g = `10ms^(-2)`).

Explanation

Consider the motion of 2kg block. The forces on it are
(i) gravitational force, `2g=2xx10=20N`. Vertically downwards,
(ii) normal reaction N by the 5kg block, vertically upwards,
(iii) force of friction `f = mu N` to the left and (iv) applied force 14 N.

In the vertical direction, there is no acceleration.
`therefore N = 20 N`
In the horizontal direction, the acceleration of the 2kg block is a.
`therefore 14-mu N = 2a " "` (`because` Resultant force = ma)
`14-0.10xx20=2a(mu=0.10)`
`14-2=2a , a = 6 ms^(-2)`.
Consider the motion of 5kg block. The forces on it are (i) gravitational force `5g=5xx10=50N`, vertically downwards,
(ii) normal reaction N of the horizontal surface, vertically upwards
(iii) force of friction `f = mu` N to the right by Newton.s third law of motion and
(iv) normal reaction, N downwards by 2kg block.
In the vertical direction, there is no acceleration.
`therefore N = 50 N`.
In the horizontal direction, the acceleration of the 5kg block is .a..
`mu N = 5a`.
`rArr 0.10 xx20=5a. , a. = 0.4 ms^(-2)`.