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When on the ground, Ian's weight is measured to be 640 N. When Ian is on an elevator, his apparent weight is \(700 \mathrm{N}\). What is the net force on the system (Ian and the elevator) if their combined mass is $1050 \mathrm{kg} ?$

Short Answer

Expert verified
Answer: The net force acting on the system (Ian and the elevator) is approximately 966 N.

Step by step solution

01

1. Determine the acceleration of the elevator

In order to determine the acceleration of the elevator, we need to find the difference in the force experienced by Ian. To do this, we will use the formula: \(\Delta F = m \cdot \Delta a\). Since we know the weight of Ian on the ground is \(640 N\), and his apparent weight on the elevator is \(700 N\), the difference in force (\(\Delta F\)) is \(700 - 640 = 60 N\). Now, let's find the mass of Ian. We know that the weight is given by the formula \(W = m \cdot g\), where \(g\) represents the gravitational acceleration (approximately \(9.8 m/s^2\)). We can rearrange this formula to find the mass: \(m = \frac{W}{g}\). Hence, Ian's mass is \(\frac{640}{9.8} \approx 65.31 kg\) Now, we can find the change in acceleration, \(\Delta a\), using the formula \(\Delta F = m \cdot \Delta a\). Rearranging we get \(\Delta a = \frac{\Delta F}{m} = \frac{60}{65.31} \approx 0.92 m/s^2\). Now that we have the acceleration of the elevator, we can move on to the next step.
02

2. Calculate the net force on the system

Now that we have the acceleration of the elevator, we can find the net force on the whole system (Ian and the elevator). The system experiences a net force given by Newton's second law, which states that the net force is equal to the mass of the system multiplied by the acceleration: \(F_{net} = m_{total} \cdot a\). We are given the combined mass of Ian and the elevator as \(1050 kg\), and we calculated the acceleration to be approximately \(0.92 m/s^2\). Therefore, we can find the net force on the system: \(F_{net} = 1050 \cdot 0.92 \approx 966 N\) So, the net force acting on the system (Ian and the elevator) is approximately \(966 N\).

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Most popular questions from this chapter

While an elevator of mass 832 kg moves downward, the tension in the supporting cable is a constant \(7730 \mathrm{N}\) Between \(t=0\) and \(t=4.00 \mathrm{s},\) the elevator's displacement is \(5.00 \mathrm{m}\) downward. What is the elevator's speed at \(t=4.00 \mathrm{s} ?\)
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