Answer:
Explanation:
Given
mass of crate=150 kg
time taken=12 s
coefficient of static friction [tex]\mu _=0.655 [/tex]
velocity of truck[tex]=50 km/h\approx 13.89 m/s[/tex]
deceleration truck experience while braking
v=u+at
[tex]0=13.89+a\times 12[/tex]
[tex]a=-1.15 m/s^2[/tex]
maximum friction force truck can offer to block
[tex]=\mu mg=0.655\times 150\times 9.8=962.85 N[/tex]
actual Force experience by truck ma[tex]=150\times 1.15=172.5[/tex]
Therefore crate will not slide during the braking
(b)range of time
[tex]a_max=6.419[/tex]
therefore
v=u+at
[tex]0=13.89-6.419\times t[/tex]
[tex]t=0.928\approx 0.93 s[/tex]
For time greater than 0.93 s there will be no sliding