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Chapter 4: Problem 21

A bag of potatoes with weight \(39.2 \mathrm{N}\) is suspended from a string that exerts a force of \(46.8 \mathrm{N}\). If the bag's acceleration is upward at \(1.90 \mathrm{m} / \mathrm{s}^{2},\) what is the mass of the potatoes?

Short Answer

Expert verified
Answer: The mass of the bag of potatoes is 4.0 kg.

Step by step solution

01

Identify the variables

In this problem, we have the following variables: 1. Weight: \(F_{weight} = 39.2 \, \mathrm{N}\) 2. Tension(force exerted by the string): \(F_{tension} = 46.8 \, \mathrm{N}\) 3. Acceleration (upward): \(a = 1.90 \, \mathrm{m} / \mathrm{s}^{2}\) 4. Mass of potatoes(m): Unknown, which is what we need to find.
02

Applying Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to its mass times its acceleration (\(F_{net} = m \times a\)). Since the problem specifies that the bag's acceleration is upward, we will assume that the upward acceleration to be positive. Thus, we can write the net force equation as follows: \(F_{net} = F_{tension} - F_{weight} = m \times a\)
03

Solve for the mass m

Now, we can plug in the given values for weight, tension, and acceleration into the equation and solve for the mass. \(46.8 \, \mathrm{N} - 39.2 \, \mathrm{N} = m \times 1.90 \, \mathrm{m} / \mathrm{s}^{2}\) \(7.6 \, \mathrm{N} = m \times 1.90 \, \mathrm{m} / \mathrm{s}^{2}\) Now, we can solve for m by dividing both sides by the acceleration value: \(m = \frac{7.6 \, \mathrm{N}}{1.90 \, \mathrm{m} / \mathrm{s}^{2}}\) \(m = 4.0 \, \mathrm{kg}\)
04

Conclusion

The mass of the bag of potatoes is 4.0 kg.

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Most popular questions from this chapter

A crate of potatoes of mass \(18.0 \mathrm{kg}\) is on a ramp with angle of incline \(30^{\circ}\) to the horizontal. The coefficients of friction are \(\mu_{\mathrm{s}}=0.75\) and \(\mu_{\mathrm{k}}=0.40 .\) Find the frictional force (magnitude and direction) on the crate if the crate is sliding up the ramp.
At what altitude above the Earth's surface would your weight be half of what it is at the Earth's surface?
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