An elevator in a building can carry a maximum of 10 persons with the average mass of each person bei

Q: Why do we add the frictional force to the weight instead of subtracting it?

Friction always opposes the direction of motion. Since the elevator is moving upwards, the frictional force acts downwards. Therefore, the motor pulling upwards must overcome both the downward weight and the downward friction, meaning we add them together.

Q: Why didn't we use the formula $F = ma$ to find the force?

We actually did! Newton's Second Law is $F_{\text{net}} = ma$. However, the problem states the elevator moves at a 'constant speed', which means acceleration ($a$) is zero. Therefore, $F_{\text{net}} = 0$, which leads to the upward forces perfectly balancing the downward forces.

Q: What if the elevator was moving downwards at a constant speed?

If the elevator were moving downwards, friction would act upwards (opposing the downward motion). The motor (or braking system) would then need to provide an upward force equal to $W - f$ to maintain a constant speed.

Q: Why use $P = Fv$ instead of $P = W/t$?

Both formulas are correct, but $P = Fv$ is much more direct here. We don't have the total work done ($W$) or the time ($t$), but we do have the instantaneous velocity ($v$). The formula $P = Fv$ is derived directly from $P = W/t = (F \times d)/t = F \times (d/t) = Fv$.

Q: Does the mass of the elevator cable matter in this calculation?

In a real-world engineering scenario, absolutely! The cable's mass adds to the total weight. However, in standard physics problems like this, if the cable's mass isn't given, we assume it is negligible or already factored into the elevator's mass.

Question

An elevator in a building can carry a maximum of $10$ persons with the average mass of each person being $68	ext{ kg}$. The mass of the elevator itself is $920	ext{ kg}$ and it moves with a constant speed of $3	ext{ m/s}$. The frictional force opposing the motion is $6000	ext{ N}$. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator ($g = 10	ext{ m/s}^2$) must be at least

Options:

(A) $62360	ext{ W}$

(B) $48000	ext{ W}$

(C) $56300	ext{ W}$

(D) $66000	ext{ W}$

Sigma Percentile · Accepted Answer

Imagine you are standing inside an elevator packed with ten people, steadily moving upwards. Have you ever wondered how much power the motor pulling you up is actually delivering? This problem is a classic application of Newton's Laws of Motion combined with the concept of Power. Let's break it down step by step.

### Analyzing the Forces

To find the power delivered by the motor, we first need to determine the exact force the motor must exert on the elevator cable. The key phrase in the problem is **"constant speed"**.

According to Newton's First Law, if an object is moving at a constant velocity, its acceleration is zero. Consequently, the net force acting on the object must also be zero. This means that the upward force provided by the motor ($F_{	ext{motor}}$) must perfectly balance all the downward forces acting on the elevator.

What are these downward forces? 
1. The total weight of the elevator and its passengers ($W_{	ext{total}}$).
2. The frictional force opposing the upward motion ($f$).

Therefore, our master force equation is:
$$F_{	ext{motor}} = W_{	ext{total}} + f$$

### Calculating the Total Weight

Let's calculate the total downward weight. We have the mass of the empty elevator ($M = 920	ext{ kg}$) and the mass of $10$ passengers. Since each person averages $68	ext{ kg}$, the total mass of the passengers is:
$$m_{	ext{passengers}} = 10 	imes 68 = 680	ext{ kg}$$

Adding this to the elevator's mass gives us the total mass of the system:
$$M_{	ext{total}} = 920 + 680 = 1600	ext{ kg}$$

To find the weight, we multiply the total mass by the acceleration due to gravity ($g = 10	ext{ m/s}^2$):
$$W_{	ext{total}} = 1600 	imes 10 = 16000	ext{ N}$$

### Determining the Motor Force

Now, the motor doesn't just have to lift this massive weight; it also has to fight against friction. The problem states that the frictional force opposing the motion is $6000	ext{ N}$. Since the elevator is moving up, friction acts downwards, adding to the burden of the motor.

Substituting our values into the force equation:
$$F_{	ext{motor}} = 16000 + 6000 = 22000	ext{ N}$$

### The Power Equation

We now have the constant force exerted by the motor ($22000	ext{ N}$) and the constant velocity of the elevator ($v = 3	ext{ m/s}$). The power ($P$) delivered by a constant force is given by the dot product of the force vector and the velocity vector:
$$P = \vec{F} \cdot \vec{v}$$

Since the motor pulls upwards and the elevator moves upwards, the angle between the force and velocity vectors is $0^\circ$. Thus, the equation simplifies to:
$$P = F 	imes v$$

### Final Calculation

Let's plug in our numbers to find the final answer:
$$P = 22000 	imes 3 = 66000	ext{ W}$$

The motor must deliver at least **$66000	ext{ W}$** of power to keep the fully loaded elevator moving upwards at a constant speed of $3	ext{ m/s}$. This matches option (d).

Always remember to check if the velocity is constant in such problems. If the elevator were accelerating, we would have to add an extra $ma$ term to our force equation to account for inertia!

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Visualized Solution

$Visualizing the System$

$Newton’s First Law$

$Calculating Total Weight$

$Total Weight Execution$

$Total Motor Force$

$Power Formula$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Kinetic Energy, Potential Energy and Power

Analyzing the Forces

Calculating the Total Weight

Determining the Motor Force

The Power Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s First Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Total Weight

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Weight Execution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Motor Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Power Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward