A 100-kg mass is sliding along a horizontal frictionless surface with a speed of 40 m/s . The mass hits a section of the surface that has friction. The coefficient of kinetic friction between the surface and the block is 0.35. How far into the friction area does the mass travel before it stops?

To determine how far the mass will travel before it stops, we need to analyze the forces acting on it.

First, let's calculate the force of kinetic friction between the surface and the block. The formula to calculate the force of kinetic friction is:

Force of Kinetic Friction = coefficient of kinetic friction x Normal Force

The normal force is equal to the weight of the object, which can be calculated using the formula:

Normal Force = mass x gravity

where mass is the mass of the block (100 kg) and gravity is the acceleration due to gravity (approximately 9.8 m/s^2).

Normal Force = 100 kg x 9.8 m/s^2 = 980 N

Next, we can calculate the force of kinetic friction:

Force of Kinetic Friction = 0.35 x 980 N = 343 N

Since there are no other horizontal forces acting on the block, the force of kinetic friction will cause the mass to decelerate. The deceleration can be calculated using Newton's second law:

Force = mass x acceleration

Rearranging the equation, we can solve for acceleration:

Acceleration = Force / mass

Acceleration = 343 N / 100 kg = 3.43 m/s^2

Now, we can calculate the time it takes for the block to stop. We'll use the equation of motion:

v = u + at

where
v = final velocity (0 m/s since the mass stops)
u = initial velocity (40 m/s)
a = acceleration (-3.43 m/s^2, negative sign indicates deceleration)
t = time

Substituting the known values into the equation:

0 = 40 m/s + (-3.43 m/s^2) x t

Solving for t:

-40 m/s = -3.43 m/s^2 x t

t = -40 m/s / (-3.43 m/s^2) ≈ 11.65 seconds

Finally, we can calculate the distance traveled during this time using the equation of motion:

s = ut + (1/2)at^2

where s = distance, u = initial velocity, t = time, and a = acceleration.

Substituting the values:

s = (40 m/s)(11.65 s) + (1/2)(-3.43 m/s^2)(11.65 s)^2

s ≈ 466 meters

Therefore, the mass will travel approximately 466 meters into the friction area before it stops.