InterviewSolution
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InterviewSolution
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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 14 N is applied to the upper block (g=10 ms^(-2)). |
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Answer» Solution :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)`. |
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