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"An ice skater applies a horizontal force to a

20.-kilogram block on frictionless, level ice,
causing the block to accelerate uniformly at
1.4 meters per second2 to the right. After the
skater stops pushing the block, it slides onto a
region of ice that is covered with a thin layer
of sand. The coefficient of kinetic friction
between the block and the sand-covered ice is
0.28.

Determine the magnitude of the normal force
acting on the block.

User Tristate
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1 Answer

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Using the Newton's Secound Law, which comes:


F_(at)=F_(r) \\ Nu=ma \\ N= (20*1.4)/(0.28) \\ \boxed {N=100N}

If you notice any mistake in my english, please let me know, because i am not native.
User Jitendra Singh
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