Answer :
Answer:
[tex]F_b=153918\, lb.ft.s^{-2}[/tex]
Explanation:
Given:
mass, [tex]m=2700 \,lb[/tex]
time, [tex]t=0.06\,s[/tex]
velocity, [tex]v=4\,ft.s^{-1}[/tex]
coefficient of kinetic friction between wheels & pavement, [tex]\mu=0.3[/tex]
According to first condition,
[tex]F\times \Delta t=m\times v[/tex]
[tex]F\times 0.06= 2700\times 4[/tex]
[tex]F=180000\,lb.ft.s^{-2}[/tex]
According to second condition,
Magnitude of frictional force (which acts opposite to the direction of motion):
[tex]f=\mu.N[/tex]
where N is the normal reaction.
[tex]f=0.3\times 2700\times 32.2[/tex]
[tex]f=26082 \,lb.ft.s^{-2}[/tex]
Now, the impulsive force on the wall if the brakes were applied during the crash:
[tex]F_b= F-f[/tex]
[tex]F_b=180000-26082[/tex]
[tex]F_b=153918\, lb.ft.s^{-2}[/tex]