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Chapter 6: Problem 67

(a) What power is needed to push a 95 -kg crate at \(0.62 \mathrm{m} / \mathrm{s}\) along a horizontal floor where the coefficient of friction is \(0.78 ?\) (b) How much work is done in pushing the crate \(11 \mathrm{m} ?\)

Short Answer

Expert verified
The power required to push the crate is approximately 450.37W, and the work done in pushing the crate 11m is 7993.93J.

Step by step solution

01

Calculate the force

First calculate the weight of the crate using the formula \(Weight= mass * g\), wherein \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)). Substituting given values, \(Weight= 95 * 9.81 = 931.95\, N\).\nNext, determine the frictional force, being the product of the weight and the coefficient of friction. Hence, \(Friction= 0.78 * 931.95 = 726.721\, N\).
02

Calculate the power

For power calculation, we use the given formula \(Power=P=Fv\). Substituting the frictional force and the crate's velocity, \(Power= 726.721 * 0.62 = 450.367 \mathrm{W}\).
03

Calculate the work done

Work done is calculated by multiplying the force (or in this case, friction) by the distance. From the given data, \(Work= Force * distance = 726.721 * 11 = 7993.93\, J\).

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Most popular questions from this chapter

A force given by \(F=b / \sqrt{x}\) acts in the \(x\) -direction, where \(b\) is a constant with the units \(\mathrm{N} \cdot \mathrm{m}^{1 / 2} .\) Show that even though the force becomes arbitrarily large as \(x\) approaches zero, the work done in moving from \(x_{1}\) to \(x_{2}\) remains finite even as \(x_{1}\) approaches zero. Find an expression for that work in the limit \(x_{1} \rightarrow 0\)

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