(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.
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