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Chapter 9: Problem 54

You're a production engineer in a cookie factory, where mounds of dough drop vertically onto a conveyer belt at the rate of one 12-g mound every 2 seconds. You're asked to design a mechanism that will keep the conveyor belt moving at a constant \(50 \mathrm{cm} / \mathrm{s} .\) What average force must the mechanism exert on the belt?

Short Answer

Expert verified
The mechanism must exert an average force of \(0.003 N\) on the belt in order to keep it moving at a constant speed of \(50 cm/s\).

Step by step solution

01

Identify Given Values

From the exercise, the given values are the mass of each mound (\(12g\)) which is \(.012 kg\), the rate at which the mounds drop onto the belt (every 2 seconds), and the constant speed needed for the belt (\(50 cm/s\)) which is \(0.5 m/s\).
02

Calculate Momentum

Momentum is the product of an object's mass and its velocity. Considering one mound every 2 seconds, we calculate momentum (\(p\)) using the provided formula: \(p = m \cdot v \), where \(m = .012 kg\) and \(v = 0.5 m/s\). Therefore, \(p = .012 kg \cdot 0.5 m/s = .006 kg \cdot m/s\).
03

Calculate the Force

The Force exerted (\(F\)) by each mound can be calculated using the formula: \(F = \Delta P/ \Delta T\), where \(\Delta P\) is the change in momentum and \(\Delta T\) is the change in time. Here \(\Delta P = .006 kg \cdot m/s\) (from step 2) and \(\Delta T = 2 s\) (as given in the problem). Therefore, \(F = .006 kg \cdot m/s / 2s = .003 N\).

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