/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A horizontal force of \(150 \mat... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 6

A horizontal force of \(150 \mathrm{~N}\) is used to push a \(40.0-\mathrm{kg}\) packing crate a distance of \(6.00 \mathrm{~m}\) on a rough horizontal surface. If the crate moves at constant speed, find (a) the work done by the \(150-\mathrm{N}\) force and (b) the coefficient of kinetic friction between the crate and surface.

Short Answer

Expert verified
The work done by the 150 N force is 900 J, and the coefficient of kinetic friction between the crate and surface is approximately 0.38.

Step by step solution

01

Calculation of Work Done

The formula for work is given by the relation \( W = F \times d \times \cos(\theta) \), where W is the work done, F is the force applied, d is the distance moved, and \( \theta \) is the angle. Since the force is in the same direction as the movement, the angle \( \theta = 0 \). Substituting the given values into the formula \( W = 150 \, \text{N} \times 6 \, \text{m} \times \cos(0) \) we get the work done.
02

Calculation of Frictional Force

Since the crate moves at a constant speed, the horizontal force is equal to the frictional force. This implies that the magnitude of the frictional force \( F_{friction} = 150 \, \text{N} \).
03

Calculation of Coefficient of Kinetic Friction

The frictional force is given by the relation \( F_{friction} = \mu_{k} \times m \times g \), where \( \mu_{k} \) is the coefficient of kinetic friction, m is the mass, and g is the gravitational constant. We can rearrange the formula to solve for the coefficient of kinetic friction as \( \mu_{k} = F_{friction} / (m \times g) \). Substituting the given values and taking \( g = 9.81 \, \text{m/s}^2 \), we will get the coefficient of kinetic friction.

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Most popular questions from this chapter

\(Q \mid C\) (a) A child slides down a water slide at an amusement park from an initial height \(h\). The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-ofenergy equation be modified? (e) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is \(12.0 \mathrm{~m}\).

In 1990 Walter Arfeuille of Belgium lifted a \(281.5-\mathrm{kg}\) object through a distance of \(17.1 \mathrm{~cm}\) using only his teeth. (a) How much work did Arfeuille do on the object? (b) What magnitude force did he exert on the object during the lift, assuming the force was constant?

S Symbolic Version of Problem 75 A light spring with spring constant \(k_{1}\) hangs from an elevated support. From its lower end hangs a second light spring, which has spring constant \(k_{2} .\) An object of mass \(m\) hangs at rest from the lower end of the second spring. (a) Find the total extension distance \(x\) of the pair of springs in terms of the two displacements \(x_{1}\) and \(x_{2}\). (b) Find the effective spring constant of the pair of springs as a system. We describe these springs as being in series.

GP A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m}\). A block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Figure \(5.20\). The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through equilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (c) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

A \(50-\mathrm{kg}\) pole vaulter running at \(10 \mathrm{~m} / \mathrm{s}\) vaults over the bar. Her speed when she is above the bar is \(1.0 \mathrm{~m} / \mathrm{s}\). Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
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