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 = 2 xx 10 = 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 14N.

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- muN = 2a` (`therefore` Resultant force = ma)
`14- 0.10 xx 20 = 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 = 5 xx 10 = 50N`, vertically downwards,
(ii) normal reaction N of the horizontal surface, vertically upwards
(iii) force of friction f = 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 xx 20 = 5a., a. = 0.4 ms^(-2)`