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Chapter 6: Q.31 (page 155)

A 50,000kg locomotive is traveling at 10m/swhen its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? Assume the track is level.

Short Answer

Expert verified

The locomotive roll about2.6×103mbefore it stops.

Step by step solution

01

Content Introduction

Deceleration is described as a reduction in speed as the body advances away from the beginning position. Negative acceleration is another name for deceleration.

02

Content Explanation

The net deceleration is provided with the formula from the equilibrium of forces F=maμsmg=maa=μsg

The rolling friction between the road and the rubber isμs=0.002

Here we are given, vi=0,vf=10

vf2+vi2=2ass=(10)22×0.002×9.81s=2.6×103m

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Most popular questions from this chapter

Very small objects, such as dust particles, experience a linear drag force,F→duag=(bv, direction opposite the motion), where bis a constant. That is, the quadratic model of drag of Equation 6.15fails for very small particles. For a sphere of radius R, the drag constant can be shown to be b=6πηR, where ηis the viscosity of the gas.

a. Find an expression for the terminal speedvtermof a spherical particle of radius Rand mass mfalling through a gas of viscosity η.

b. Suppose a gust of wind has carried50−μma -diameter dust particle to a height of 300m. If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density ofrole="math" localid="1647776411884" 2700kg/m3the viscosity of25∘Cair is 2.0×10−5Ns/m2, and you can assume that the falling dust particle reaches terminal speed almost instantly.

A 10kg crate is placed on a horizontal conveyor belt. The materials are such that μs= 0.5and μk= 0.3.

a. Draw a free-body diagram showing all the forces on the crate if the conveyor belt runs at constant speed.

b. Draw a free-body diagram showing all the forces on the crate if the conveyor belt is speeding up.

c. What is the maximum acceleration the belt can have without the crate slipping?
  1. Write a realistic problem for which these are the correct equations.
  2. Draw the free-body diagram and the pictorial representation for your problem.
  3. Finish the solution of the problem.
(100N)cos⁡30∘−fk=(20kg)ax

n+(100N)sin⁡30∘−(20kg)(9.80m/s2)=0fk=0.20n.

The forces in FIGURE EX6.8act on a 2.0kgobject. What are the values of axand ay, the x- and y-components of the object’s acceleration?

Are the following statements true or false? Explain.

a. The mass of an object depends on its location.

b. The weight of an object depends on its location.

c. Mass and weight describe the same thing in different units.
See all solutions

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