## Understanding Power Requirements for an Elevator System### ConceptualizeIn this scenario, the motor must supply a force of magnitude \( T \) to pull the elevator car upward with increasing speed. More power is required than in part (A) due to the added task of accelerating the car.### CategorizeThe elevator car is modeled as a **particle under a net force**, since it is accelerating.### AnalyzeApply **Newton's second law** to the car:\[\sum F_y = T - f - Mg = Ma\]**Solve for \( T \):**\[T = M(a + g) + f\]Using the **power formula** \( P = \vec{F} \cdot \vec{v} \), the required power becomes:\[P = Tv = [M(a + g) + f]v\]**Substitute numerical values:**\[P = [(1,680 \, \text{kg})(1.80 \, \text{m/s}^2 + 9.80 \, \text{m/s}^2) + 4,050 \, \text{N}]v\]\[P = \underline{\hspace{2cm}} \, v\]---### FinalizeTo compare with part (A), let \( v = 2.20 \, \text{m/s} \), resulting in power:\[P = T(2.20 \, \text{m/s}) = \underline{\hspace{2cm}} \, \text{W}\]This expectedly results in larger power than part (A).---### Master ItIf the same \( 1,460 \, \text{kg} \) elevator with a \( 220 \, \text{kg} \) load descends at \( 2.20 \, \text{m/s} \), what power is delivered by the motor? (Consider energy removed to slow descent.)\[\underline{\hspace{2cm}} \, \text{W}\] **Example 8.11: Power Delivered by an Elevator Motor**An elevator car (see figure (a)) has a mass of 1,460 kg and is carrying passengers with 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 2.20 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 1.80 m/s²?---**SOLVE IT****(A)** How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 2.20 m/s?**Conceptualize**: The motor must supply the force of magnitude \( T \) that pulls the elevator car upward.**Categorize**: The friction force increases the power necessary to lift the elevator. The problem states that the speed of the elevator is constant, which tells us that \( a = 0 \). We model the elevator as a particle in equilibrium.**Analyze**: The free-body diagram in figure (b) specifies the upward direction as positive. The total mass \( M \) of the system (car plus passengers) is equal to 1,680 kg.Using the particle in equilibrium model, apply Newton’s second law to the car:\[\sum F_y = T - f - Mg = 0\]Solve for \( T \):\[T = Mg + f\]Use \( P = \vec{F} \cdot \vec{v} \) and that \( \vec{T} \) is in the same direction as \( \vec{v} \) to find the power:\[P = \vec{T} \cdot \vec{v} = Tv = (Mg + f)v\]Substitute numerical values:\[P = [(1,680 \, \text{kg})(9.80 \, \text{m/s}^2) + (4,050 \, \text{N})](2.20 \, \text{m/s}) = \, \text{W}\]---**(B)** What power must the motor deliver at the

Mass of the vehicle, m1=1460 Kg Mass of the people inside it, m2=220 Kg Total mass,…

Conceptualize In this case, the motor must supply the force of magnitude T that pulls the eleva upward with an increasing speed. We expect that more power will be required to do that than in 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 EFy =T-f- Mg = Ma car: Solve for T: T = M(a + g) + f Use P = F. v to obtain the required power: P = Tv = [M(a + g) + f]v P = [(1,680 kg)(1.80 m/s² + 9.80 m/s²) · Substitute numerical values: Finalize To compare with part (A), let v = 2.20 m/s, giving a power of P = T(2.20 m/s) = w which is larger than the power found in part (A), as expected. MASTER IT Suppose the same (1,460 kg) elevator with the same (220 kg) load descends at 2.20 m/s. What power delivered by the motor? (Energy must be removed to slow the elevators descent.) w

Conceptualize In this case, the motor must supply the force of magnitude T that pulls the eleva upward with an increasing speed. We expect that more power will be required to do that than in 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 EFy =T-f- Mg = Ma car: Solve for T: T = M(a + g) + f Use P = F. v to obtain the required power: P = Tv = [M(a + g) + f]v P = [(1,680 kg)(1.80 m/s² + 9.80 m/s²) · Substitute numerical values: Finalize To compare with part (A), let v = 2.20 m/s, giving a power of P = T(2.20 m/s) = w which is larger than the power found in part (A), as expected. MASTER IT Suppose the same (1,460 kg) elevator with the same (220 kg) load descends at 2.20 m/s. What power delivered by the motor? (Energy must be removed to slow the elevators descent.) w