Question

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?

Answer 1

Answer:

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 :

$v^2-u^2=2ad$

$-u^2=2ad$

$a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2$

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

F = ma

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

|F| = 2 N

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