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?

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

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 :

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

F = ma

|F| = 2 N

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

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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