Answer to Question #102793 in Civil and Environmental Engineering for John

Question

A 2 tonnes car is towed by a 2.5 tonnes truck which moves off from rest and reaches a
speed of 2m/s before the tow-rope tightens on the stationary car. The rope then stretches
taking up the strain for 0.15 seconds.Find:
(a) the common speed when the rope ceases to stretch
(b) the average impulsive force in the rope.(Neglect resistance and the effect of the truck’s
propelling force during the stretching period

Answer 1

a)force=mass*acceleration

=(2.5*1000+2*1000)*2m/s

F=9000N

Which is R_t =9000N

a_t=R_t/m=9000N/2000

=4.5m/s

now the common speed is 4.5m/s

b)impulse force =force*time

=9000N*0.15second

=1350N.S