M3W6 Problem Set #6-Singlar

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Mechanical Engineering

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Dec 6, 2023

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Katelyn Singlar PHYS1200 M3W6 Problem Set #6 Problem Set #6-- Problems 2, 10, 18, 26, 48 2. Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver 2? The stopping distance of driver 1 will be greater than the stopping distance of driver 2. (b) Choose the best explanation from among the following: I. Locking up the brakes gives the greatest possible braking force. II. The same tires on the same road result in the same force of friction. III. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction. Option III (Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.) is the best choice to explain why driver 1 will take longer to stop than driver 2. 10. A person places a cup of coffee on the roof of his car while he dashes back into the house for a forgotten item. When he returns to the car, he hops in and takes off with the coffee cup still on the roof. (a) If the coefficient of static friction between the coffee cup and the roof of the car is 0.24, what is the maximum acceleration the car can have without causing the cup to slide? Ignore the effects of air resistance. Given: µ=0.24 g=9.81m/s2 Unknown: a=? Equation: a=µg Substitution: a=(0.24)(9.81m/s 2 ) Solution: a=2.3544m/s 2 Solution (2 sig figs): a=2.4m/s 2 (b) What is the smallest amount of time in which the person can accelerate the car from rest to 15 m/s and still keep the coffee cup on the roof? Given: Vf=15m/s Vi=0m/s a=2.4m/s 2 Unknown: t=? Equation: t=change in velocity/a Substitution: t=(15m/s-0m/s)/2.4m/s 2 Solution: t=6.25s Solution (2 sig figs): t=6.3s

18. A backpack full of books weighing 52.0 N rests on a table in a physics laboratory classroom. A spring with a force constant of 150 N/m is attached to the backpack and pulled horizontally. (a) If the spring is pulled until it stretches 2.00 cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? Given: k=150N/m X=.0200m Unknown: Ff=? Equation: F=-kx Substitution: F=-(150N/m)(.0200m) Solution: F=-3N Solution (2 sig figs): F=-3.0N (b) Does your answer to part (a) change if the mass of the backpack is doubled? Explain. Doubling the mass of the backpack does not change the answer to part a because the force constant on the spring is unaffected by the mass of the backpack. In the problem as presented in part a, the spring did not pull the backpack, so doubling the mass of the backpack would not change anything in the problem and the force of friction would be the same. 26. After a skiing accident, your leg is in a cast and supported in a traction device as shown in Figure -43. Find the magnitude of the force exerted by the leg on the small pulley. (By Newton’s third law, the small pulley exerts an equal and opposite force on the leg.) Let the mass m be 2.27. Given: m=2.27kg g=9.81m/s 2 Angle=30 degrees Unknown: F=? Equation: F=2mg(cos(angle)) Substitution: F=2(2.27kg)(9.81m/s 2 )(cos30) Solution: F=38.5705N Solution (2 sig figs): F=39N 48. To test the effects of high acceleration on the human body, the National Aeronautics and Space Administration (NASA) constructed a large centrifuge at the Manned Spacecraft Center in Houston. In this device, astronauts were placed in a capsule that moved in a circular path with a radius of 15 m. If the astronauts in this centrifuge experienced a centripetal acceleration 9.0 times that of gravity, what was the linear speed of the capsule? Given: a=88.29m/s2 (9*9.81) r=15m Unknown: v=? Equation: v=square root (ra) Substitution: v=square root ((15m)(88.29m/s2) Solution: v=36.3916199m/s Solution (2 sig figs): v=36m/s

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