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6) Bubba is driving along one day with a box in the bed of his truck....

6) Bubba is driving along one day with a box in the bed of his truck. He makes a right turn into a parking lot. The radius of the turn is 21.5 feet. The coefficient of friction between the box and the bed of the truck is 0.177. What is the fastest speed that Bubba can turn into the parking lot without the box sliding in the bed of the truck?

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Answer #1

Radius of turn is. r= 21.5 feet

Coefficient of friction between the box and the bed of the truck is =0.177

Now the bubba performing the circular motion

Hence

Maximum speed of the bubba so that box couldn't slide in the bed can be written as

Now 1feet=30.48 cm

r=21.5*30.48 = 655.32 cm = 6.5532m

Now the acceleration due to gravity=g = 9.8 m/s

This is the maximum velocity

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