A 1200 kg car travels at a constant speed of 15 m/s in a circular loop with a diameter of 45 m. What is the coefficient of friction on the car?

(a)

KE = m v^2 / 2 = (1200 kg)(20 m/s)^2 / 2 = 240,000 J

(b)

The energy is entirely dissipated by the force of friction in the brake system.

(c)

W = delta KE = KEf - KEi = (0 - 240,000) J = -240,000 J

(d)

Fd = delta KE

F = (delta KE) / d = (-240,000 J) / (50 m) = -4800 N

The magnitude of the friction force is 4800 N.

mark brainliest if correct

i believe

step-by-step explanation:

from my conclusion c

take the logarithm of both sides (the base of the log doesn't matter):

apply the exponent property - this says that :

distribute the log terms on both sides:

group like terms together:

divide through both sides by the coefficient on :

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like: