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Two identical blocks are stacked on top of each other and placed on a table. To overcome the force of static friction, a force of 10 N is required. If the blocks were placed side by side and pushed as shown in the figure above, how much force would be required to move them?(A) $\frac{10 \sqrt{2}}{2 \mathrm{N}}$(B) 10 $\mathrm{N}$(C) 10$\sqrt{2} \mathrm{N}$(D) 2 $\mathrm{ON}$

B

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Chapter 1

Practice Test 1

Section 1

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here we have to find out new fourth required to move to overcome the fourth of static friction when the two blocks are change from position of top of on top of each other, two side by side. So when they are on top of each other, the normal force is equal to mg weight of this system. What other system is combined? Weight after two blocks and when they are side by side, the normal forces again weight of the system. So the normal force is combined weight of two blocks. Since friction force depends on coefficient of kinetic friction and normal force, coefficient of kinetic fixing does not change. A normal force does not change in both to cases. So the static friction fourth limits same and the force required to move them will also remain the same. So the correct option is B 10 Newtons, the required forced to move them. This completes the solution. Thank you

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