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A 2.0 kg block, initially moving at 10.0 m/s, slides 50.0 m across a sheet of ice beforecoming to rest. What is the magnitude of the average frictional force on the block?

User ShyJ
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1 Answer

2 votes

Answer:

The magnitude of the average frictional force on the block is 2 N.

Step-by-step explanation:

Given that.

Mass of the block, m = 2 kg

Initial velocity of the block, u = 10 m/s

Distance, d = 50 m

Finally, it stops, v = 0

Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :


v^2-u^2=2ad


-u^2=2ad


a=(-u^2)/(2d)\\\\a=(-(10\ m/s)^2)/(2* 50\ m)\\\\a=-1\ m/s^2

The frictional force on the block is given by the formula as :

F = ma


F=2\ kg* (-1)\ m/s^2\\\\F=-2\ N

|F| = 2 N

So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.

User Vexe
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