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Chapter 11: Problem 4

A worker pushed a \(58.7\) -lb block \((m=26.6 \mathrm{~kg})\) a distance of \(31.3 \mathrm{ft}(=9.54 \mathrm{~m})\) along a level floor at constant speed with a force directed \(32.0^{\circ}\) below the horizontal. The coefficient of kinetic friction is \(0.21\). How much work did the worker do on the block?

Short Answer

Expert verified
The worker did 1163.959 work on the block, in units of lb \cdot ft.

Step by step solution

01

Calculate the horizontal component of the force.

This can be done using the formula for the x-component of a force, which is given by \( F_{x} = F \cdot \cos(\Theta) \). Here, \( F \) is the magnitude of the force and \( \Theta \) is the angle the force makes with the horizontal direction. Plugging in the values, we get \( F_{x} = 58.7 \lb \cdot \cos(32.0^{\circ}) = 49.53 \lb \).
02

Calculate the force of friction.

The force of friction can be calculated using the formula \( F_{f} = \mu \cdot F_{N} \), where \( \mu \) is the coefficient of friction and \( F_{N} \) is the normal force. In this situation, since the block is moving along a level surface, the normal force is just the weight of the block. Therefore, \( F_{f} = 0.21 \cdot 58.7 \lb = 12.327 \lb \).
03

Calculate the net force applied by the worker.

Since the block moves at a constant speed, the horizontal component of the force applied by the worker is equal to the force of friction. So, the net force \( F_{net} = F_{x} - F_{f} = 49.53 \lb - 12.327 \lb = 37.203 \lb \).
04

Calculate the work done.

The work done by a force is given by \( W = F_{net} \cdot d \), where \( d \) is the distance traveled. Substituting the given values, we find that \( W = 37.203 \lb \cdot 31.3 ft = 1163.959 lb \cdot ft \).

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Most popular questions from this chapter

What power is developed by a grinding machine whose wheel has a radius of \(20.7 \mathrm{~cm}\) and runs at \(2.53\) rev/s when the tool to be sharpened is held against the wheel with a force of \(180 \mathrm{~N}\) ? The coefficient of friction between the tool and the wheel is \(0.32\)

\((a)\) Calculate \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\), where \(\overrightarrow{\mathbf{a}}=5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\), \(\overrightarrow{\mathbf{b}}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\), and \(\overrightarrow{\mathbf{c}}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} .\) (b) Calculate the angle between \(\overrightarrow{\mathbf{r}}\) and the \(+z\) axis. ( \(c\) ) Find the angle between \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\).

A swimmer moves through the water at a speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The drag force opposing this motion is \(110 \mathrm{~N}\). How much power is developed by the swimmer?

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