A 25,000 kilogram train is traveling down a track at 20 meters per second. A cow wanders onto the tracks 75 meters ahead of the train, causing the conductor to slam on the brakes. The rain skids to a stop. If the brakes can provide 62,500 Newtons of friction, will the conductor have enough stopping distance to avoid striking the cow?

1/2(25,000)(20M/S)^2= 500,000/62500n= 80 j

Is this correct? Thank you

To determine whether the train will have enough stopping distance to avoid striking the cow, we can use the concept of kinetic friction and the equation of motion.

First, let's calculate the initial kinetic energy (KE) of the train using the given mass (m) and velocity (v):

KE = 1/2 * m * v^2

Plugging in the values:

KE = 1/2 * 25,000 kg * (20 m/s)^2
= 1/2 * 25,000 kg * 400 m^2/s^2
= 5,000,000 kg * m^2/s^2
= 5,000,000 J

The initial kinetic energy of the train is 5,000,000 Joules.

Next, let's analyze the braking force provided by the friction. The maximum frictional force (F) that can be exerted by the brakes is given as 62,500 N.

Now, we need to calculate the work (W) done by the braking force to bring the train to a stop. The work is equal to the change in kinetic energy (W = ΔKE) of the train, which, in this case, is equal to the initial kinetic energy (KE) because the train comes to a stop.

W = ΔKE = KE

Since the work is equal to the force (F) applied multiplied by the stopping distance (d), we can rewrite the equation as:

W = F * d

Plugging in the values:

62,500 N * d = 5,000,000 J

To find the stopping distance (d), we divide both sides by the force (62,500 N):

d = 5,000,000 J / 62,500 N
= 80 m

Therefore, the stopping distance required for the train to avoid striking the cow is 80 meters.

In conclusion, based on the given maximum frictional force and the distance between the cow and the train, the conductor does have enough stopping distance (80 meters) to avoid striking the cow.