1a) i) Tractive force = 700 N, ii) Tension =340N
Consider as a whole the tractor and trailer first and then consider the trailer to find the tension. The friction is 100N and 400 N respectively .
bi) Tractive force 2200N ii) Tension = 441N
You have to include the weight component mgsin(angle) to the system as a whole and the system is has net force =0. Consider the trailer to find tension.
ci) a = 1.7ms-2 , t= 11.5s, v = 11.5 ms-1
The acceleration of the trailer is gsin(angle) if the tension becomes zero, down the slope. Applying equation of motion, to avoid quadratic equation , find final velocity v first. Taking up as -ve , u= -8 ms-1, a = 1.7ms-2 and s = 20m , v is found to be 11.5 ms-1. Then t can be found using v = u + at , no forgeting to use u = -8 ms -1 .
2i) F=153.3N
ii) a= 1.7 ms-2
The acceleration due to weight component only is gsin(angle) =1.7 ms-2.
iii) F braking = 153N
The braking force = weight component.
iv) Heat energy
Loss in potential energy is converted to heat energy in the brakes)
Rate of work done against braking force = Fv (same as power)
v) power = 859W
3. Tension in cable= 28.1 N
Tension in spring= 22.8 N
x = 0.0285 N
Use component method to solve for the two tensions, then use F=kx to find extension
