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Chapter 5: Problem 49

A 310 -g paperback book rests on a 1.2 -kg textbook. A force is applied to the textbook, and the two books accelerate together from rest to \(96 \mathrm{cm} / \mathrm{s}\) in \(0.42 \mathrm{s}\). The textbook is then brought to a stop in \(0.33 \mathrm{s}\), during which time the paperback slides off. Within what range does the coefficient of static friction between the two books lie?

Short Answer

Expert verified
The coefficient of static friction between the two books lies in the range from 0 to 0.23.

Step by step solution

01

Determine the Acceleration

From the problem, we know the initial velocity \( u = 0 \) cm/s, final velocity \( v = 96 \) cm/s and time \( t = 0.42 \) s. We convert the velocity from cm/s to m/s, \( v = 0.96 \) m/s. Now we can use the equation of motion \( v = u + at \) to find the acceleration \( a \), where \( u \) is the initial velocity, \( v \) is the final velocity, \( a \) is the acceleration, and \( t \) is the time. So, \( a = (v - u) / t = (0.96 - 0) / 0.42 = 2.29 \) m/s².
02

Calculate the Total Force

Next calculate the total force \( F \) using the formula \( F = ma \), where \( m \) is the total mass of the two books (1.2 kg for the textbook and 0.31 kg for the paperback book converting from grams to kg) and \( a \) is acceleration. So, \( F = (1.2 + 0.31) * 2.29 = 3.47 \) N.
03

Compute the Maximum Frictional Force

The maximum frictional force \( F_{max} \) can be calculated using Newton's second law \( F_{max} = ma \) with the mass \( m \) of the top book (paperback book) and the acceleration \( a \) from the first step. So, \( F_{max} = 0.31 * 2.29 = 0.71 \) N.
04

Calculate Normal Force

The normal force \( N \) is equal to the weight of the top book (paperback book). The weight of an object is its mass times the acceleration due to gravity \( g = 9.8 \) m/s². So, \( N = mg = 0.31 * 9.8 = 3.04 \) N.
05

Determine the Coefficient of Static Friction

Finally, to get the coefficient of static friction \( \mu_{s} \), use the formula \( \mu_{s} = F_{max} / N \). So, \( \mu_{s} = 0.71 / 3.04 = 0.23 \).
06

Conclusions

Because static friction takes values from zero up to a maximum value, and we have calculated the maximum value, the coefficient of static friction \( \mu_{s} \) can take values from 0 to 0.23.

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Most popular questions from this chapter

You're investigating a subway accident in which a train derailed while rounding an unbanked curve of radius \(132 \mathrm{m},\) and you're asked to estimate whether the train exceeded the \(45-\mathrm{km} / \mathrm{h}\) speed limit for this curve. You interview a passenger who had been standing and holding onto a strap; she noticed that an unused strap was hanging at about a \(15^{\circ}\) angle to the vertical just before the accident. What do you conclude?

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