Answer:
The question implies that force of friction is 2.1E4 N
F = M v^2 / R frictional force needed to supply this centripetal force
2.1E4 = M v^2 / R
v^2 = R * 2.1E4 / M
v^2 = 120 * 2.1E4 / 2.5E3 = 1200 * 2.1 / 2.5 = 1008 m^2/s^2
v = 31.7 m/s max speed of truck (about 70 MPH)
