An elevator car (see figure (a)) has a mass of 1,560 kg and is carrying passenge

Question

An elevator car (see figure (a)) has a mass of 1,560 kg and is carrying passengers having a combined mass of 220 kg. A constant friction force of 4,050 N retards its motion.

(A) How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 2.00 m/s?

(B) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1.70 m/s2?

Please show full steps for all parts especially A and B, I'm struggling to understand the calculations.

(A) How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 2.00 m/s? M g Conceptualize The motor must supply the force of magnitude that pulls the elevator car upward. Categorize The friction force increases the power necessary (a) The matar exerts an upward force T onthe to lift the elevator. The problem states that the speed of the elevator car. The magnitude Tin the cables connecting the car elevator is constant, which tells us that a 0, We model the and motor. The downward forces acting elevator as a particle in equilibrium. car are a friction force and the gravitational force Mg. (b) The free-body diagram for Analyze The free-body diagram in figure (b specifies the upward direction as positive. The total mass Mof the system (car plus passengers) is equal to 1,780 kg Using the particle in equilibrium model, apply f- Mig 30 Newton's second law to the car: Solve for T. Use P F v and that T is in the same direction as v to find the power P 1,780 kg)(9.80 m/s (4,050 N)](2.00 m/s) Substitute numerical values: 5544 (B) What power must the motor deliver at the instant the speed of the elevator is v ifthe motor is designed to provide the elevator car with an upward acceleration of 1.70 m/s Conceptualize In this case, the motor must supply the force of magnitude Tthat pulls the elevator car upward with an increasing speed. We expect that more power will be required to do that than in part (A) because the motor must now perform the additional task of accelerating the car. Categorize In this case, we model the elevator car as a particle under a net force because it is accelerating. Analyze Apply Newton's second law to the F, Ms J Me Solve for T. f 2 g) fl F v to obtain the required power 2 Use P P 1,780 k 1.70 m/ 9.80 m/s 4,050 N1v Substitute numerical values: Finalize To compare with part (A), let v 2.00 m/s, giving a power of P T2.00 m/s) which is larger than the power found in part (A), as expected. MASTER IT Suppose the same (1,560 kg) elevator with the same 220 kg) load descends at 2.00 m/s. hat is the power delivered by the motor? nergy must be removed to slow the elevators descent.

Explanation / Answer

a) T = M ( a+g) + f = ( 1560 + 220 ) kg ( 0 +9.8) + 4050 ( a= 0 , because elevator is moving at constant velocity)

T = 21494 N

power = Tv =21494 (2) = 42988 watts

b) T = ( 1560 + 220 ) kg ( 1.70 +9.8) + 4050= 1780 ( 11.5) + 4050 = 24520 N

power = 24520 v watts= 49040 watts

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