A 9230-kg truck collides head on with a 1250-kg parked car. The vehicles entangle together and slide a linear distance of 10.6 meters before coming to rest. Assuming a uniform coefficient of friction of 0.820 between the road surface and the vehicles, determine the pre-collision speed of the truck.

=>By the work energy relation:-

=>W = ∆KE

=>Ff.s = 1/2(M+m)v^2

=>µk x N x s = 1/2(M+m)v^2

=>µk x (M+m) x g x s = 1/2(M+m)v^2

=>v = √[2µkgs]

=>v = √[2 x 0.82 x 9.8 x 10.6]

=>v = 13.05 m/s

By the law of momentum conservation:-

=>Mu = (M+m)v

=>9230u = (9230+1250) x 13.05

=>u = 14.82 m/s

Step-by-step explanation:

step-by-step explanation:

sry but any answer choices