Cutnell, Physics, 10e signment PRINTER VERSION BACK NEXT Chapter 04, Problem X23

Question

Cutnell, Physics, 10e signment PRINTER VERSION BACK NEXT Chapter 04, Problem X23 X Incorrect. Consult 4.2 Interactive LearningWare before beginning this problem. A truck is traveling at a speed of 23.5 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient of static friction between the crate and the truck bed is 0.623. Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck. 3.96 the tolerance is +/-5% Question Attempts: 2 of 5 used SAVE FOR LATER SUBMIT ANSWER Version 4.24.1 vacy Policy I 9 2000-2017 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley& Sons.Ins

Explanation / Answer

Let:
m = mass of the crate
a = the deceleration
Vf = the final velocity = 0
Vi = the initial velocity = 23.5m/s
t = the shortest time to go from 23.5m/s to 0
s = the distance traveled in time t
s = the coefficient of static friction = 0.623
g = the acceleration of gravity = 9.81m/s^2

The maximum frictional force Ff between the crate and the bed of the truck is:
Ff = s * m * g

The deceleration force Fd is given by:
Fd = m * a

This cannot exceed the frictional force; so we have:
m * a = Ff = s * m * g

Canceling the m, we get:
a = s * g

From our EoM (#(6)), we know that:
a = (Vf - Vi) / t

= (0 - Vi) / t = -Vi / t

So,
-Vi / t = s * g
t = -Vi / (s * g)

= -23.5 / (0.623 * -9.81)

= 3.84s = the shortest amount of time the truck can take to stop.

Using another EoM we can calculate the distance it will cover during the deceleration:
s = (1/2) * a * t^2

= (1/2) * (Vi / t) * t^2

= (1/2) * Vi * t

= (1/2) * 23.5 * 3.845

= 45.2 m

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