Problem 5.32 - enhanced - with feedback the rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. suppose a 180,000 kg locomotive is rolling at 22 m/s on level rails. you may want to review ( pages 138 - 142) . part a if the engineer disengages the engine, how much time will it take the locomotive to coast to a stop?

The frictional force can be calculate as:

Ff = μr * N

where μr is the frictional constant while N is the normal force which is also equivalent to weight, hence:

Ff = 0.002 * 180,000 kg * 9.81 m/s^2

Ff = 3,531.6 N

The frictional force is also equivalent to the product of mass and acceleration, so we can find a:

Ff = m * a

a = 3,531.6 N / 180,000 kg

a = 0.01962 m/s^2 (in negative direction)

We can solve for time using the formula:

v = vi + a t

where v is final velocity = 0, vi is initial velocity = 22 m/s, t is time

0 = 22 - 0.01962 * t

t = 1,121.3 seconds

each chemical family is represented as a group or column on the periodic table. based on chemical properties, there are 6 types of chemical families. group 1 represents alkali metals, group 2 represents alkaline earth metals, groups 3 to 12 represent transition metals, groups 13 to 16 represent non-metals and other metals, group 17 represents halogens and group 18 represents noble gases.