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An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg.

An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg.

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2. INSTRUCTIONS: - Show the 'free body diagrams' completely. - Show all equations obtained from the free body diagrams, and how equations are derived from the FBD. - Show all solutions completely. An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg. It takes a force of 53 kN on the lead car of a 16-car passenger train with mass 9.1 x 105 kg to pull it at a constant 45 m/s (101 mi/h) on level tracks. (a) What power must the locomotive provide to the lead car? (b) How much more power to the lead car than calculated in part (a) would be needed to give the train an acceleration of 1.5 m/s², at the instant that the train has a speed of 45 m/s on level tracks? (c) How much more power to the lead car than that calculated in part (a) would be needed to move the train up a 1.5% grade (slope angle a = arctan 0.015) at a constant 45 m/s? Riding a Loop-the-Loop. A car in an amusement park ride rolls without friction around a track. The car starts from rest at point A at a heighth above the bottom of the loop. Treat the car as a particle. (a) What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)? (b) If h = 3.50R and R = 14.0 m, compute the speed and radial acceleration at point C at which is halfway down to the bottom of the circular path? Hint: Solve for minimum speed at the top of the loop. Use net force F = mg=mv^2/R where normal force N = 0 at minimum speed. h B COR
Transcribed Image Text:2. INSTRUCTIONS: - Show the 'free body diagrams' completely. - Show all equations obtained from the free body diagrams, and how equations are derived from the FBD. - Show all solutions completely. An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg. It takes a force of 53 kN on the lead car of a 16-car passenger train with mass 9.1 x 105 kg to pull it at a constant 45 m/s (101 mi/h) on level tracks. (a) What power must the locomotive provide to the lead car? (b) How much more power to the lead car than calculated in part (a) would be needed to give the train an acceleration of 1.5 m/s², at the instant that the train has a speed of 45 m/s on level tracks? (c) How much more power to the lead car than that calculated in part (a) would be needed to move the train up a 1.5% grade (slope angle a = arctan 0.015) at a constant 45 m/s? Riding a Loop-the-Loop. A car in an amusement park ride rolls without friction around a track. The car starts from rest at point A at a heighth above the bottom of the loop. Treat the car as a particle. (a) What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)? (b) If h = 3.50R and R = 14.0 m, compute the speed and radial acceleration at point C at which is halfway down to the bottom of the circular path? Hint: Solve for minimum speed at the top of the loop. Use net force F = mg=mv^2/R where normal force N = 0 at minimum speed. h B COR
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