in the end mass will not matter
initial energy = (1/2) m v^2 = 50 m
friction force = .4 m g
work done = force * distance = .4 m g * 2 = .8 m g
final Ke = 50 m - .8 m g = m (50-.8 g)
so if u = final speed
(1/2) m u^2 = m (50 - .8 g)
u^2 = 100 - 1.6 g
-----------------------------------
force * distance to stop = (1/2) m v^2 = 50 m
.4 m g d = 50 m
d = 50/(.4g)
Can someone explain to me how I can solve this problem? This problem contains multiple parts but I'm stuck on the last two. Here is the full problem: A block of mass 2.5 kg is sliding at 10 m/s on an initially frictionless, horizontal surface. There is then a 2-m patch of roughness where the friction is present (u=0.4), beyond which the surface is once again frictionless.
1. How fast is the block moving on the far side of the rough patch?
2. What would the length of the rough patch of the surface have to be to bring the block to a complete stop?
1 answer