A car of mass 1200kg pulls a trailer of mass of 400kg along a straight horizontal road. The car and trailer are connected by a tow-rope modelled as a light inextensible rod. The engine of the car provides a constant driving force of 3200N. (see ? below)

a. Since the car-trailer combo is accelerating at 0.4 m/s^2, the net force is

#F_"net" = m*a = (1200 kg+400 kg)*0.4 m/s^2 = 640 N#

The remainder of the 3200N is dealing with the resistances to motion. So friction, air drag, etc. of both vehicles add up to #3200N - 640N = 2560N#. That was ignoring the direction of this force. These resistances are forces pointing to the rear. Therefore we need to give this 2560 N a negative sign.

Because the horizontal resistances of the car and the trailer are proportional to their respective masses, the trailer's resistance, #R_t#, is

#R_t = -2560N*(400 cancel(kg))/(1200 cancel(kg)+400 cancel(kg)) = -640 N#

b. The tension, #T# in the tow rope is providing force to deal with #R_t# and also to provide the acceleration. Note that this is a force pointing forward -- it is encouraging the trailer to keep up with the car.

#T = 640 N + ma = 640 N + 400 kg0.4 m/s^2 = 800 N

c. After the tow rope breaks, #R_t# will eventually stop it (assuming it stays on the road). #R_t# is a force of 640 N and that will provide acceleration #a_t# of (solving Newton's 2nd Law for #a_t#

#a_t = R_t/m = (-640 N)/(400 kg) = -1.6 m/s^2#

#v^2 = u^2 + 2*a_t*d#

#0^2 = (25ms^-1)^2 + 2*(-1.6 m/s^2)*d#

#-2*(-1.6 m/s^2)*d = (25ms^-1)^2#

#d = (625 (mcancel(^2))/cancel(s^2))/(3.2 cancel(m)/cancel(s^2)) = 2000 m#

I hope this helps,
Steve