a 1200 kg trailer is hitched to a 1400 kg car. the car and trailer are traveling at 72 km.h when the driver applies the brakes on both the car and the trailer. knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N respectively, determine (a) the distance traveled by the car and trailer before they come to a stop and (b) the horizontal component of the force exerted by the trailer hitch

Respuesta :

Answer:

a) 8.67m

b) 1000N

Explanation:

(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:

[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]

once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:

[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]

v: final velocity=0

vo: initial velocity = 72km/h = 60 m/s

by replacing the values of these parameters you obtain for x:

[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]

(b) The horizontal component of the force exerted by the trailer hitch is given by:

[tex]F_T=5000N-4000N=1000N[/tex]