The friction force causes the block to decelerate. I will use the following equation to determine the distance.

vf^2 = vi^2 + 2 * a * d, vf = 0

Ff = -0.25 * m * 9.8

a = - 2.45 m/s^2

0 = 10^2 + 2 * -2.45 * d

d = 100 ÷ 4.9

The distance is approximately 20 meters.

It will stop when all the KE is used up by work. So KE = 1/2 mU^2 = kmgS = QE work. And S = U^2/2kg = 100/(2*.25*9.8) = 20.41 m. ANS.

The frictional force is 0.25*mg, so the acceleration is -0.25g, or -2.45 m/s^2.

2ax = (vf)^2 - (v0)^2 =>

2(-2.45 m/s^2)* x = -100 m^2/s^2.

x = (100/4.9) m = about 20 m but use a calculator.